13. Multiple and Submultiple Angles Solutions

  1. Let us solve these one by one.

    1. Given, cos⁡A=35\cos A = \frac{3}{5}

      ⇒sin⁡A=1−cos⁡2A=1−925=1625=45\Rightarrow \sin A = \sqrt{1 - \cos^2A} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}

      sin⁡2A=2sin⁡Acos⁡A=2.45.35=2425\sin 2A = 2\sin A\cos A = 2.\frac{4}{5}.\frac{3}{5} = \frac{24}{25}

    2. Given, sin⁡A=1213\sin A = \frac{12}{13}

      ⇒cos⁡A=1−sin⁡2A=1−144169=25169=513\Rightarrow \cos A = \sqrt{1 - \sin^2A} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}

      sin⁡2A=2sin⁡Acos⁡A=2.1213.513=120169\sin 2A = 2\sin A\cos A = 2.\frac{12}{13}.\frac{5}{13} = \frac{120}{169}

    3. Given, tan⁡A=1663=perpendicularbase\tan A = \frac{16}{63} = \frac{\text{perpendicular}}{\text{base}}

      hypotenuse=p2+b2=162+632=65\text{hypotenuse} = \sqrt{p^2 + b^2} = \sqrt{16^2 + 63^2} = 65

      sin⁡A=1665,cos⁡A=6365\sin A = \frac{16}{65}, \cos A = \frac{63}{65}

      sin⁡2A=2sin⁡Acos⁡A=2.1665.6365=20164225\sin 2A = 2\sin A\cos A = 2.\frac{16}{65}.\frac{63}{65} = \frac{2016}{4225}

  2. Let us solve these one by one.

    1. Given, cos⁡A=1517\cos A = \frac{15}{17}

      ⇒sin⁡A=1−cos⁡2A\Rightarrow \sin A = \sqrt{1 - \cos^2A} =1−225289=64289=817= \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{64}{289}} = \frac{8}{17}

      cos⁡2A=cos⁡2A−sin⁡2A=225−64289=161289\cos 2A = \cos^2A - \sin^2A = \frac{225 - 64}{289} = \frac{161}{289}

    2. Given, sin⁡A=45\sin A = \frac{4}{5}

      ⇒cos⁡A=1−sin⁡2A\Rightarrow \cos A = \sqrt{1 - \sin^2A} =1−1625=35= \sqrt{1 - \frac{16}{25}} = \frac{3}{5}

      cos⁡2A=cos⁡2A−sin⁡2A=9−1625=−725\cos2A = \cos^2A - \sin^2A = \frac{9 - 16}{25} = -\frac{7}{25}

    3. Give, tan⁡A=512=perpendicularbase\tan A = \frac{5}{12} = \frac{\text{perpendicular}}{\text{base}}

      hypotenuse=p2+b2=25+144=13\text{hypotenuse} = \sqrt{p^2 + b^2} = \sqrt{25 + 144} = 13

      sin⁡A=513,cos⁡A=1213\sin A = \frac{5}{13}, \cos A = \frac{12}{13}

      cos⁡2A=cos⁡2A−sin⁡2A=119169\cos^2A = \cos^2A - \sin^2A = \frac{119}{169}

  3. Given, tan⁡A=ba,\tan A = \frac{b}{a}, thus hypotenuse=b2+a2\text{hypotenuse} = \sqrt{b^2 + a^2}

    acos⁡2A+bsin⁡2A=a(cos⁡2A−sin⁡2A)+2bsin⁡Acos⁡Aa\cos 2A+ b\sin 2A = a(\cos^2A - \sin^2A) + 2b\sin A\cos A

    =a(a2a2+b2−b2a2+b2)+2b.aba2+b2= a\left(\frac{a^2}{a^2 + b^2} - \frac{b^2}{a^2 + b^2}\right) + 2b.\frac{ab}{a^2 + b^2}

    =a(a2−b2+2b2a2+b2)=a= a\left(\frac{a^2 - b^2 + 2b^2}{a^2 + b^2}\right) = a

  4. We have to prove that sin⁡2A1+cos⁡2A=tan⁡A\frac{\sin 2A}{1 + \cos 2A} = \tan A

    L.H.S. =sin⁡2A1+cos⁡2A=2sin⁡Acos⁡A1+cos⁡2A−sin⁡2A= \frac{\sin 2A}{1 + \cos 2A} = \frac{2\sin A\cos A}{1 + \cos^2A - \sin^2A}

    =2sin⁡Acos⁡A2cos⁡2A[∵1−sin⁡2A=cos⁡2A]= \frac{2\sin A\cos A}{2\cos^2A}[\because 1 - \sin^2A = \cos^2A]

    =tan⁡A== \tan A = R.H.S.

  5. We have to prove that sin⁡2A1−cos⁡2A=cot⁡A\frac{\sin 2A}{1 - \cos 2A} = \cot A

    L.H.S. =sin⁡2A1−cos⁡2A=2sin⁡Acos⁡A1−(cos⁡2A−sin⁡2A)= \frac{\sin 2A}{1 - \cos 2A} = \frac{2\sin A\cos A}{1 -(\cos^2A - \sin^2A)}

    =2sin⁡Acos⁡A2sin⁡2A=cot⁡A== \frac{2\sin A\cos A}{2\sin^2A} = \cot A = R.H.S.

  6. We have to prove that 1−cos⁡2A1+cos⁡2A=tan⁡2A\frac{1 - \cos 2A}{1 + \cos 2A} = \tan^2A

    L.H.S. =1−(cos⁡2A−sin⁡2A)1+cos⁡2A−sin⁡2A= \frac{1 - (\cos^2A - \sin^2A)}{1 + \cos^2A - \sin^2A}

    =2sin⁡2A2cos⁡2A=tan⁡2A== \frac{2\sin^2A}{2\cos^2A} = \tan^2A = R.H.S.

  7. We have to prove that tan⁡A+cot⁡A=2cosec⁡2A\tan A + \cot A = 2\cosec 2A

    L.H.S. =sin⁡Acos⁡A+cos⁡Asin⁡A=sin⁡2A+cos⁡2Asin⁡Acos⁡A= \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2A + \cos^2A}{\sin A\cos A}

    =22sin⁡Acos⁡A=2sin⁡2A=2cosec⁡2A== \frac{2}{2\sin A\cos A} = \frac{2}{\sin 2A} = 2\cosec 2A = R.H.S.

  8. We have to prove that tan⁡A−cot⁡A=−2cot⁡2A\tan A - \cot A = -2\cot2A

    L.H.S. =sin⁡Acos⁡A−cos⁡Asin⁡A=sin⁡2A−cos⁡2Asin⁡Acos⁡A= \frac{\sin A}{\cos A} - \frac{\cos A}{\sin A} = \frac{\sin^2A - \cos^2A}{\sin A\cos A}

    =−cos⁡2Asin⁡2A2=−2cot⁡2A== \frac{-\cos2A}{\frac{\sin2A}{2}} = -2\cot2A = R.H.S.

  9. We have to prove that cosec⁡2A+cot⁡2A=cot⁡A\cosec 2A + \cot 2A = \cot A

    L.H.S. =1sin⁡2A+cos⁡2Asin⁡2A=1+cos⁡2Asin⁡2A=2cos⁡2A2sin⁡Acos⁡A= \frac{1}{\sin 2A} + \frac{\cos 2A}{\sin 2A} = \frac{1 + \cos 2A}{\sin 2A} = \frac{2\cos^2A}{2\sin A\cos A}

    =cot⁡A== \cot A = R.H.S.

  10. We have to prove that 1−cos⁡A+cos⁡B−cos⁡(A+B)1+cos⁡A−cos⁡B−cos⁡(A+B)=tan⁡A2cot⁡B2\frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)} = \tan\frac{A}{2}\cot\frac{B}{2}

    L.H.S. =1−cos⁡A+cos⁡B−cos⁡(A+B)1+cos⁡A−cos⁡B−cos⁡(A+B)= \frac{1 - \cos A + \cos B - \cos(A + B)}{1 + \cos A - \cos B - \cos(A + B)}

    =2sin⁡2A2+2sin⁡A2sin⁡(A2+B)2cos⁡2A2−2cos⁡A2cos⁡(A2+B)= \frac{2\sin^2\frac{A}{2} + 2\sin\frac{A}{2}\sin\left(\frac{A}{2} + B\right)}{2\cos^2\frac{A}{2} - 2\cos\frac{A}{2}\cos\left(\frac{A}{2} + B\right)}

    =sin⁡A2(sin⁡A2+sin⁡(A2+B))cos⁡A2(cos⁡A2−cos⁡(A2+B))= \frac{\sin\frac{A}{2}\left(\sin\frac{A}{2} + \sin \left(\frac{A}{2} + B\right)\right)}{\cos\frac{A}{2}\left(\cos\frac{A}{2} - \cos\left(\frac{A}{2} + B\right)\right)}

    =tan⁡A2(2sin⁡(A+B2)cos⁡B2)2sin⁡(A+B2)sin⁡B2= \frac{\tan\frac{A}{2}\left(2\sin\left(\frac{A + B}{2}\right)\cos\frac{B}{2}\right)}{2\sin\left(\frac{A + B}{2}\right)\sin\frac{B}{2}}

    =tan⁡A2cot⁡B2= \tan\frac{A}{2}\cot\frac{B}{2}

  11. We have to prove that cos⁡A1∓sin⁡A=tan⁡(45∘±A2)\frac{\cos A}{1 \mp \sin A} = \tan\left(45^\circ \pm \frac{A}{2}\right)

    First considering −- sign on L.H.S.,

    L.H.S. =cos⁡A1−sin⁡A=cos⁡2A2−sin⁡2A2(cos⁡A2−sin⁡A2)2= \frac{\cos A}{1 - \sin A} = \frac{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}{\left(\cos\frac{A}{2} - \sin\frac{A}{2}\right)^2}

    Dividing numerator and denomiator by cos⁡2A2\cos^2\frac{A}{2}

    =1−tan⁡2A2(1−tan⁡A2)2= \frac{1 - \tan^2\frac{A}{2}}{\left(1 - \tan\frac{A}{2}\right)^2}

    =1+tan⁡A21−tan⁡A2= \frac{1 + \tan\frac{A}{2}}{1 - \tan\frac{A}{2}}

    =tan⁡45∘+tan⁡A21−tan⁡45∘tan⁡A2=tan⁡(45∘+A2)= \frac{\tan45^\circ + \tan\frac{A}{2}}{1 - \tan45^\circ\tan\frac{A}{2}} = \tan\left(45^\circ + \frac{A}{2}\right)

    Similarly by considering the ++ sign we can prove the other sign.

  12. We have to prove that sec⁡8A−1sec⁡4A−1=tan⁡8Atan⁡2A\frac{\sec 8A - 1}{\sec 4A - 1} = \frac{\tan 8A}{\tan 2A}

    L.H.S. =sec⁡8A−1sec⁡4A−1=1−cos⁡8A1−cos⁡4A.cos⁡4Acos⁡8A= \frac{\sec 8A - 1}{\sec 4A - 1} = \frac{1 - \cos 8A}{1 - \cos 4A}.\frac{\cos4A}{\cos8A}

    =2sin⁡24A2sin⁡22A.cos⁡4Acos⁡8A=(2sin⁡4Acos⁡4A).sin⁡4A2sin⁡22A.cos⁡8A= \frac{2\sin^24A}{2\sin^22A}.\frac{\cos 4A}{\cos8A} = \frac{(2\sin4A\cos4A).\sin4A}{2\sin^22A.\cos8A}

    =sin⁡8Acos⁡8A.sin⁡4A2sin⁡22A=tan⁡8A.2sin⁡2Acos⁡2A2sin⁡22A=tan⁡8Atan⁡2A== \frac{\sin8A}{\cos8A}.\frac{\sin4A}{2\sin^22A} = \frac{\tan8A. 2\sin2A\cos2A}{2\sin^22A} = \frac{\tan8A}{\tan2A} = R.H.S.

  13. We have to prove that 1+tan⁡2(45∘−A)1−tan⁡2(45∘−A)=cosec⁡2A\frac{1 + \tan^2(45^\circ - A)}{1 - \tan^2(45^\circ - A)} = \cosec 2A

    L.H.S. =1+tan⁡2(45∘−A)1−tan⁡2(45∘−A)= \frac{1 + \tan^2(45^\circ - A)}{1 - \tan^2(45^\circ - A)}

    =cos⁡2(45∘−A)+sin⁡2(45∘−A)cos⁡2(45∘−A)−sin⁡2(45∘−A)= \frac{\cos^2(45^\circ - A) + \sin^2(45^\circ - A)}{\cos^2(45^\circ - A) - \sin^2(45^\circ - A)}

    =1cos⁡(90∘−2A)=1sin⁡2A=cosec⁡2A== \frac{1}{\cos(90^\circ - 2A)} = \frac{1}{\sin2A} = \cosec2A = R.H.S.

  14. We have to prove that sin⁡A+sin⁡Bsin⁡A−sin⁡B=tan⁡A+B2tan⁡A−B2\frac{\sin A + \sin B}{\sin A - \sin B} = \frac{\tan \frac{A + B}{2}}{\tan \frac{A - B}{2}}

    L.H.S. =sin⁡A+sin⁡Bsin⁡A−sin⁡B=2sin⁡A+B2cos⁡A−B22cos⁡A+B2sin⁡A−B2= \frac{\sin A + \sin B}{\sin A - \sin B} = \frac{2\sin\frac{A + B}{2}\cos\frac{A - B}{2}}{2\cos\frac{A + B}{2}\sin\frac{A - B}{2}}

    =tan⁡A+B2tan⁡A−B2== \frac{\tan \frac{A + B}{2}}{\tan \frac{A - B}{2}} = R.H.S.

  15. We have to prove that sin⁡2A−sin⁡2Bsin⁡Acos⁡A−sin⁡Bcos⁡B=tan⁡(A+B)\frac{\sin^2A - \sin^2B}{\sin A\cos A - \sin B\cos B} = \tan(A + B)

    L.H.S. =2(cos⁡2B−cos⁡2A)sin⁡2A−sin⁡2B=cos⁡2B−cos⁡2Asin⁡2A−sin⁡2B= \frac{2(\cos^2B - \cos^2A)}{\sin2A - \sin2B} = \frac{\cos2B - \cos2A}{\sin2A - \sin2B}

    =sin⁡(A+B)sin⁡(A−B)cos⁡(A+B)sin⁡(A−B)=tan⁡(A+B)== \frac{\sin(A + B)\sin(A - B)}{\cos(A + B)\sin(A - B)} = \tan(A + B) = R.H.S.

  16. We have to prove that tan⁡(π4+A)−tan⁡(π4−A)=2tan⁡2A\tan\left(\frac{\pi}{4} + A\right) - \tan\left(\frac{\pi}{4} - A\right) = 2\tan 2A

    L.H.S. =1+tan⁡A1−tan⁡A−1−tan⁡A1+tan⁡A= \frac{1 + \tan A}{1 - \tan A} - \frac{1 - \tan A}{1 + \tan A}

    =(1+tan⁡A)2−(1−tan⁡A)21−tan⁡2A=4tan⁡A1−tan⁡2A= \frac{(1 + \tan A)^2 - (1 - \tan A)^2}{1 - \tan^2A} = \frac{4\tan A}{1 - \tan^2A}

    =4sin⁡Acos⁡A.cos⁡2Acos⁡2A−sin⁡2A=2sin⁡2Acos⁡2A=2tan⁡2A== \frac{4\sin A}{\cos A}. \frac{\cos^2A}{\cos^2A - \sin^2A} = \frac{2\sin2A}{\cos2A} = 2\tan2A = R.H.S.

  17. We have to prove that cos⁡A+sin⁡Acos⁡A−sin⁡A−cos⁡A−sin⁡Acos⁡A+sin⁡A=2tan⁡2A\frac{\cos A + \sin A}{\cos A - \sin A} - \frac{\cos A - \sin A}{\cos A + \sin A} = 2\tan 2A

    L.H.S. =(cos⁡A+sin⁡A)2−(cos⁡A−sin⁡A)2cos⁡2A−sin⁡2A= \frac{(\cos A + \sin A)^2 - (\cos A - \sin A)^2}{\cos^2A - \sin^2A}

    =4cos⁡Asin⁡Acos⁡2A=2sin⁡2Acos⁡2A=2tan⁡2A== \frac{4\cos A\sin A}{\cos 2A} = \frac{2\sin 2A}{\cos 2A} = 2\tan2A = R.H.S.

  18. We have to prove that cot⁡(A+15∘)−tan⁡(A−15∘)=4cos⁡2A1+2sin⁡2A\cot (A + 15^\circ) - \tan(A - 15^\circ) = \frac{4\cos 2A}{1 + 2\sin 2A}

    L.H.S. =1−tan⁡(A+15∘)tan⁡(A−15∘)tan⁡(A+15∘)= \frac{1 - \tan(A + 15^\circ)\tan(A - 15^\circ)}{\tan(A + 15^\circ)}

    =cos⁡(A+15∘)cos⁡(A−15∘)−sin⁡(A+15∘)sin⁡(A−15∘)cos⁡(A+15∘)cos⁡(A−15∘).cos⁡(A+15∘)sin⁡(A+15∘)= \frac{\cos(A + 15^\circ)\cos(A - 15^\circ) - \sin(A + 15^\circ)\sin(A - 15^\circ)}{\cos(A + 15^\circ)\cos(A - 15^\circ)}.\frac{\cos(A + 15^\circ)}{\sin(A + 15^\circ)}

    =cos⁡2Asin⁡(A+15∘)cos⁡(A−15∘)=2cos⁡2A2sin⁡(A+15∘)cos⁡(A−15∘)= \frac{\cos 2A}{\sin(A + 15^\circ)\cos(A - 15^\circ)} = \frac{2\cos 2A}{2\sin(A + 15^\circ)\cos(A - 15^\circ)}

    =2cos⁡2Asin⁡2A+sin⁡30∘=4cos⁡2A1+sin⁡2A== \frac{2\cos 2A}{\sin2A + \sin30^\circ} = \frac{4\cos 2A}{1 + \sin 2A} = R.H.S.

  19. We have to prove that sin⁡A+sin⁡2A1+cos⁡A+cos⁡2A=tan⁡A\frac{\sin A + \sin2A}{1 + \cos A + \cos 2A} = \tan A

    L.H.S. =sin⁡A+2sin⁡Acos⁡Acos⁡A+2cos⁡2A=sin⁡A(1+2cos⁡A)cos⁡A(1+2cos⁡A)= \frac{\sin A + 2\sin A\cos A}{\cos A + 2\cos^2A} = \frac{\sin A(1 + 2\cos A)}{\cos A(1 + 2\cos A)}

    =tan⁡A== \tan A = R.H.S.

  20. We have to prove that 1+sin⁡A−cos⁡A1+sin⁡A+cosA=tan⁡A2\frac{1 + \sin A - \cos A }{1 + \sin A + cos A} = \tan \frac{A}{2}

    L.H.S. =2sin⁡2A2+2sin⁡A2cos⁡A22cos⁡2A2+2sin⁡A2cos⁡A2= \frac{2\sin^2\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}

    =sin⁡A2(sin⁡A2+cos⁡A2)cos⁡A2(sin⁡A2+cos⁡A2)= \frac{\sin\frac{A}{2}(\sin\frac{A}{2} + \cos\frac{A}{2})}{\cos\frac{A}{2}(\sin\frac{A}{2} + \cos\frac{A}{2})}

    =tan⁡A2== \tan\frac{A}{2} = R.H.S.

  21. We have to prove that sin⁡(n+1)A−sin⁡(n−1)Acos⁡(n+1)A+2cos⁡nA+cos⁡(n−1)A=tan⁡A2\frac{\sin(n + 1)A - \sin(n - 1)A}{\cos(n + 1)A + 2\cos nA + \cos(n - 1)A} = \tan \frac{A}{2}

    L.H.S. =2cos⁡nAsin⁡A2cos⁡nAcos⁡A+2cos⁡nA=sin⁡A1+cos⁡A= \frac{2\cos nA \sin A}{2\cos nA \cos A + 2\cos nA} = \frac{\sin A}{1 + \cos A}

    =2sin⁡A2cos⁡A22cos⁡2A2=tan⁡A2== \frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^2\frac{A}{2}} = \tan \frac{A}{2} = R.H.S.

  22. We have to prove that sin⁡(n+1)A+2sin⁡nA+sin⁡(n−1)Acos⁡(n−1)−cos⁡(n+1)A=cot⁡A2\frac{\sin(n + 1)A + 2\sin nA + \sin(n - 1)A}{\cos(n - 1) - \cos(n + 1)A} = \cot \frac{A}{2}

    L.H.S. =2sin⁡nAcos⁡A+2sin⁡nA2sin⁡nAsin⁡A= \frac{2\sin nA\cos A + 2\sin nA}{2\sin nA\sin A}

    =cos⁡A+1sin⁡A=2cos⁡2A22sin⁡A2cos⁡A2= \frac{\cos A + 1}{\sin A} = \frac{2\cos^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}}

    =cot⁡A2== \cot\frac{A}{2} = R.H.S.

  23. We have to prove that sin⁡(2n+1)Asin⁡A=sin⁡2(n+1)A−sin⁡2nA\sin(2n + 1)A\sin A = \sin^2(n + 1)A - \sin^2nA

    R.H.S. =(sin⁡(n+1)A+sin⁡nA)(sin⁡(n+1)A−sin⁡nA)= (\sin(n + 1)A + \sin nA)(\sin(n + 1)A - \sin nA)

    =(2sin⁡2n+12Acos⁡A2)(2cos⁡2n+12Asin⁡A2)= (2\sin\frac{2n + 1}{2}A\cos \frac{A}{2})(2\cos\frac{2n + 1}{2}A\sin \frac{A}{2})

    =2sin⁡2n+12Acos⁡2n+12A.2cos⁡A2sin⁡A2= 2\sin\frac{2n + 1}{2}A\cos\frac{2n + 1}{2}A.2\cos \frac{A}{2}\sin\frac{A}{2}

    =sin⁡(2n+1)Asin⁡A== \sin(2n + 1)A\sin A = L.H.S.

  24. We have to prove that sin⁡(A+3B)+sin⁡(3A+B)sin⁡2A+sin⁡2B=2cos⁡(A+B)\frac{\sin(A + 3B) + \sin(3A + B)}{\sin 2A + \sin 2B} = 2\cos(A + B)

    L.H.S. =sin⁡(A+3B)+sin⁡(3A+B)sin⁡2A+sin⁡2B= \frac{\sin(A + 3B) + \sin(3A + B)}{\sin 2A + \sin 2B}

    =2sin⁡(2A+2B)cos⁡(A−B)2sin⁡(A+B)cos⁡(A−B)= \frac{2\sin(2A + 2B)\cos(A - B)}{2\sin(A + B)\cos(A - B)}

    =2sin⁡(A+B)cos⁡(A+B)sin⁡(A+B)=2cos⁡(A+B)== \frac{2\sin(A + B)\cos(A + B)}{\sin(A + B)} = 2\cos(A + B) = R.H.S.

  25. We have to prove that sin⁡3A+sin⁡2A−sin⁡A=4sin⁡Acos⁡A2cos⁡3A2\sin 3A + \sin 2A - \sin A = 4\sin A\cos \frac{A}{2}\cos \frac{3A}{2}

    L.H.S. =2cos⁡2Asin⁡A+2sin⁡Acos⁡A=2sin⁡A(cos⁡2A+cos⁡A)= 2\cos 2A\sin A + 2\sin A\cos A = 2\sin A(\cos 2A + \cos A)

    =2sin⁡Acos⁡3A2cos⁡A2== 2\sin A\cos \frac{3A}{2}\cos\frac{A}{2} = R.H.S.

  26. We have to prove that tan⁡2A=(sec⁡2A+1)sec⁡2A−1\tan 2A = (\sec 2A + 1)\sqrt{\sec^2A - 1}

    R.H.S. =1+cos⁡2Acos⁡2A1−cos⁡2Acos⁡2A= \frac{1 + \cos 2A}{\cos 2A}\sqrt{\frac{1 - \cos^2A}{\cos^2A}}

    =2cos⁡2A2cos⁡2A−1.sin⁡2Acos⁡2A= \frac{2\cos^2A}{2\cos^2A - 1}.\sqrt{\frac{\sin^2A}{\cos^2A}}

    =22−sec⁡2A.tanA=2tan⁡A1−tan⁡2A=tan⁡A+tan⁡A1−tan⁡A.tan⁡A= \frac{2}{2 - \sec^2A}.tan A = \frac{2\tan A}{1 - \tan^2A} = \frac{\tan A + \tan A}{1 - \tan A.\tan A}

    =tan⁡2A==\tan 2A = R.H.S.

  27. We have to prove that cos⁡32A+3cos⁡2A=4(cos⁡6A−sin⁡6A)\cos^32A + 3\cos 2A = 4(\cos^6A - \sin^6A)

    L.H.S. =(cos⁡2A−sin⁡2A)3+3(cos⁡2A−sin⁡2A)= (\cos^2A - \sin^2A)^3 + 3(\cos^2A - \sin^2A)

    =cos⁡6A−3cos⁡4Asin⁡2A+3cos⁡2Asin⁡4A−sin⁡6A+3(cos⁡2A−sin⁡2A)= \cos^6A -3\cos^4A\sin^2A + 3\cos^2A\sin^4A - \sin^6A + 3(\cos^2A - \sin^2A)

    =cos⁡6A−3cos⁡4A(1−cos⁡2A)+3(1−sin⁡2A)sin⁡4A−sin⁡6A+3(cos⁡2A−sin⁡2A)= \cos^6A -3\cos^4A(1 - \cos^2A) + 3(1 - \sin^2A)\sin^4A - \sin^6A + 3(\cos^2A - \sin^2A)

    =4(cos⁡6A−sin⁡6A)== 4(\cos^6A - \sin^6A) = R.H.S.

  28. We have to prove that 1+cos⁡22A=2(cos⁡4A+sin⁡4A)1 + \cos^22A = 2(\cos^4A + \sin^4A)

    L.H.S. =1+(cos⁡2A−sin⁡2A)2=1−2sin⁡2Acos⁡2A+cos⁡4A+sin⁡4A= 1 + (\cos^2A - \sin^2A)^2 = 1 - 2\sin^2A\cos^2A + \cos^4A + \sin^4A

    =1−2sin⁡2A(1−sin⁡2A)+cos⁡4A+sin⁡4A= 1 - 2\sin^2A(1 - \sin^2A) + \cos^4A + \sin^4A

    =1−2sin⁡2A+2sin⁡4A+cos⁡4A+sin⁡4A= 1 - 2\sin^2A + 2\sin^4A + \cos^4A + \sin^4A

    =(1−sin⁡2A)2+cos⁡4A+2sin⁡4A=2(cos⁡4A+sin⁡4A)== (1 - \sin^2A)^2 + \cos^4A + 2\sin^4A = 2(\cos^4A + \sin^4A) = R.H.S.

  29. We have to prove that sec⁡2A(1+sec⁡2A)=2sec⁡2A\sec^2A(1 + \sec2A) = 2\sec2A

    L.H.S. =1cos⁡2A.cos⁡2A+1cos⁡2A= \frac{1}{\cos^2A}.\frac{\cos2A + 1}{\cos 2A}

    =1cos⁡2A.2cos⁡2Acos⁡2A=2sec⁡2A== \frac{1}{\cos^2A}.\frac{2\cos^2A}{\cos 2A} = 2\sec2A = R.H.S.

  30. We have to prove that cosec⁡A−2cot⁡2Acos⁡A=2sin⁡A\cosec A - 2\cot 2A\cos A = 2\sin A

    L.H.S. =1sin⁡A−2cos⁡2Acos⁡Asin⁡2A= \frac{1}{\sin A} - \frac{2\cos 2A\cos A}{\sin 2A}

    =1sin⁡A−2cos⁡2Acos⁡A2sin⁡Acos⁡A= \frac{1}{\sin A} - \frac{2\cos 2A\cos A}{2\sin A\cos A}

    1sin⁡A−cos⁡2Asin⁡A=1−cos⁡2Asin⁡A\frac{1}{\sin A} - \frac{\cos 2A}{\sin A} = \frac{1 - \cos 2A}{\sin A}

    =2sin⁡2Asin⁡A=2sin⁡A== \frac{2\sin^2A}{\sin A} = 2\sin A = R.H.S.

  31. We have to prove that cot⁡A=12(cot⁡A2−tan⁡A2)\cot A = \frac{1}{2}\left(\cot\frac{A}{2} - \tan\frac{A}{2}\right)

    R.H.S. =12(1−tan⁡2A2tan⁡A2)= \frac{1}{2}\left(\frac{1 - \tan^2\frac{A}{2}}{\tan\frac{A}{2}}\right)

    =12(cos⁡2A2−sin⁡2A2cos⁡2A2).cos⁡A2sin⁡A2= \frac{1}{2}\left(\frac{\cos^2\frac{A}{2} - \sin^2\frac{A}{2}}{\cos^2\frac{A}{2}}\right).\frac{\cos\frac{A}{2}}{\sin \frac{A}{2}}

    =12cos⁡Acos⁡A2.1sin⁡A2=cot⁡A== \frac{1}{2}\frac{\cos A}{\cos\frac{A}{2}}.\frac{1}{\sin\frac{A}{2}} = \cot A = L.H.S.

  32. We have to prove that sin⁡Asin⁡(60∘−A)sin⁡(60∘+A)=14sin⁡3A\sin A\sin(60^\circ - A)\sin(60^\circ + A) = \frac{1}{4}\sin 3A

    L.H.S. =sin⁡A.cos⁡2A−cos⁡120∘2=sin⁡A(1−2sin⁡2A+12)2=\sin A.\frac{\cos 2A - \cos 120^\circ}{2} = \frac{\sin A\left(1 - 2\sin^2A + \frac{1}{2}\right)}{2}

    =3sin⁡A−4sin⁡3A4=14sin⁡3A== \frac{3\sin A - 4\sin^3A}{4} = \frac{1}{4}\sin 3A = R.H.S.

  33. We have to prove that cos⁡Acos⁡(60∘−A)cos⁡(60∘+A)=14cos⁡3A\cos A\cos(60^\circ - A)\cos(60^\circ + A) = \frac{1}{4}\cos 3A

    L.H.S. =cos⁡A2(cos⁡2A+cos⁡120∘)=cos⁡A2(2cos⁡2A−1−12)= \frac{\cos A}{2}\left(\cos 2A + \cos120^\circ\right) = \frac{\cos A}{2}\left(2\cos^2A - 1 - \frac{1}{2}\right)

    =4cos⁡3A−3cos⁡A4=14cos⁡3A== \frac{4\cos^3A - 3\cos A}{4} = \frac{1}{4}\cos 3A = R.H.S.

  34. We have to prove that cot⁡A+cot⁡(60∘+A)−cot⁡(60∘−A)=3cot⁡3A\cot A + \cot(60^\circ + A) - \cot(60^\circ - A) = 3\cot 3A

    L.H.S. =1tan⁡A+1tan⁡(60∘+A)−1tan⁡(60∘−A)= \frac{1}{\tan A} + \frac{1}{\tan(60^\circ + A)} - \frac{1}{\tan(60^\circ - A)}

    =1tan⁡A+1−3tan⁡A3+tan⁡A−1+3tan⁡A3−tan⁡A= \frac{1}{\tan A} + \frac{1 - \sqrt{3}\tan A}{\sqrt{3} + \tan A} - \frac{1 + \sqrt{3}\tan A}{\sqrt{3} - \tan A}

    =1tan⁡A−8tan⁡A3−tan⁡2A=3(1−3tan⁡2A)3tan⁡A−tan⁡3A=3tan⁡3A= \frac{1}{\tan A} - \frac{8\tan A}{3 - \tan^2A} = \frac{3(1 - 3\tan^2A)}{3\tan A - \tan^3A} = \frac{3}{\tan 3A}

    =3cot⁡3A== 3\cot 3A = R.H.S.

  35. We have to prove that cos⁡4A=1−8cos⁡2A+8cos⁡4A\cos 4A = 1 - 8\cos^2A + 8\cos^4A

    L.H.S. =cos⁡4A=2cos⁡22A−1=2(2cos⁡2A−1)2−1= \cos 4A = 2\cos^22A - 1 = 2(2\cos^2A - 1)^2 - 1

    =2(4cos⁡4A−4cos⁡2A+1)−1=2(4\cos^4A - 4\cos^2A + 1) - 1

    =1−8cos⁡2A+8cos⁡4A== 1 - 8\cos^2A + 8\cos^4A = R.H.S.

  36. We have to prove that sin⁡4A=4sin⁡Acos⁡3A−4cos⁡Asin⁡3A\sin 4A = 4\sin A\cos^3A - 4\cos A\sin^3A

    L.H.S. =2sin⁡2Acos⁡2A=4sin⁡Acos⁡A(cos⁡2A−sin⁡2A)= 2\sin 2A\cos 2A = 4\sin A\cos A(\cos^2A - \sin^2A)

    =4sin⁡Acos⁡3A−4cos⁡Asin⁡3A== 4\sin A\cos^3A - 4\cos A\sin^3A = R.H.S.

  37. We have to prove that cos⁡6A=32cos⁡6A−48cos⁡4A+18cos⁡2A−1\cos 6A = 32\cos^6A - 48\cos^4A + 18\cos^2A - 1

    L.H.S. =cos⁡6A=(cos⁡23A−sin⁡23A)=(4cos⁡3A−3cos⁡A)2−(3sin⁡A−4sin⁡3A)2= \cos 6A = (\cos^23A - \sin^23A) = (4\cos^3A - 3\cos A)^2 - (3\sin A - 4\sin^3A)^2

    =16cos⁡6A+9cos⁡2A−24cos⁡4A−9sin⁡2A−16sin⁡6A+24sin⁡4A= 16\cos^6A + 9\cos^2A -24\cos^4A - 9\sin^2A - 16\sin^6A + 24\sin^4A

    =16cos⁡6A+9cos⁡2A−24cos⁡4A−9(1−cos⁡2A)−16(1−cos⁡2A)3+24(1−cos⁡2A)2= 16\cos^6A + 9\cos^2A -24\cos^4A - 9(1 - \cos^2A) - 16(1 - \cos^2A)^3 + 24(1 - \cos^2A)^2

    =32cos⁡6A−48cos⁡4A+18cos⁡2A−1== 32\cos^6A - 48\cos^4A + 18\cos^2A - 1 = R.H.S.

  38. We have to prove that tan⁡3Atan⁡2Atan⁡A=tan⁡3A−tan⁡2A−tan⁡A\tan 3A\tan 2A\tan A = \tan 3A - \tan 2A - \tan A

    Rewriting this as following:

    tan⁡A+tan⁡2A=tan⁡3A(1−tan⁡Atan⁡2A)⇒tan⁡A+tan⁡2A1−tan⁡Atan⁡2A=tan⁡3A\tan A + \tan 2A = \tan 3A(1 - \tan A\tan 2A)\Rightarrow \frac{\tan A + \tan 2A}{1 - \tan A\tan 2A} = \tan 3A

    ⇒tan⁡(A+2A)=tan⁡3A\Rightarrow \tan (A + 2A) = \tan 3A

    Hence, proved.

  39. We have to prove that 2cos⁡2nA+12cos⁡A+1=(2cos⁡A−1)(2cos⁡2A−1)(2cos⁡22A−1)…(2cos⁡2n−1−1)\frac{2\cos2^nA + 1}{2\cos A + 1} = (2\cos A - 1)(2\cos 2A - 1)(2\cos2^2A - 1)\ldots(2\cos2^{n - 1} - 1)

    L.H.S. =2cos⁡2nA+12cos⁡A+1= \frac{2\cos2^nA + 1}{2\cos A + 1}

    Multiplying and dividing by 2cos⁡A−12\cos A - 1

    =(2cos⁡A−1)2cos⁡2nA+14cos⁡2A−1=(2cos⁡A−1)2cos⁡2nA+12cos⁡2A+1= (2\cos A - 1)\frac{2\cos2^nA + 1}{4\cos^2A - 1} = (2\cos A - 1)\frac{2\cos2^nA + 1}{2\cos2A + 1}

    Multiplying and dividing by 2cos⁡2A−12\cos2A - 1

    =(2cos⁡A−1)(2cos⁡2A−1)2cos⁡2nA+14cos⁡22A−1= (2\cos A - 1)(2\cos2A - 1)\frac{2\cos2^nA + 1}{4\cos^22A - 1}

    =(2cos⁡A−1)(2cos⁡2A−1)2cos⁡2nA+12cos⁡22A+1= (2\cos A - 1)(2\cos2A - 1)\frac{2\cos2^nA + 1}{2\cos2^2A + 1}

    Proceeding similarly we obtain the R.H.S.

  40. Given tan⁡A=17,sin⁡B=110\tan A= \frac{1}{7}, \sin B = \frac{1}{\sqrt{10}}

    ∴cos⁡B=310,tan⁡B=13\therefore \cos B = \frac{3}{\sqrt{10}}, \tan B = \frac{1}{3}

    tan⁡(A+2B)=tan⁡A+tan⁡2B1−tan⁡Atan⁡2B\tan(A + 2B) = \frac{\tan A + \tan 2B}{1 - \tan A\tan 2B}

    =tan⁡A+2tan⁡B1−tan⁡2B1−tan⁡A.2tan⁡B1−tan⁡2B= \frac{\tan A + \frac{2\tan B}{1 - \tan^2B}}{1 - \tan A.\frac{2\tan B}{1 - \tan^2B}}

    =17+2131−191−17.2131−19= \frac{\frac{1}{7} + \frac{2\frac{1}{3}}{1 - \frac{1}{9}}}{1 - \frac{1}{7}.\frac{2\frac{1}{3}}{1 - \frac{1}{9}}}

    =1∴A+2B=π4= 1 \therefore A + 2B = \frac{\pi}{4}

  41. We have to prove that tan⁡(π4+A)+tan⁡(π4−A)=2sec⁡2A\tan\left(\frac{\pi}{4} + A\right) + \tan\left(\frac{\pi}{4} - A\right) = 2\sec2A

    L.H.S. =1+tan⁡A1−tan⁡A+1−tan⁡A1+tan⁡A= \frac{1 + \tan A}{1 - \tan A} + \frac{1 - \tan A}{1 + \tan A}

    =(1+tan⁡A)2+(1−tan⁡A)21−tan⁡2A=2+2tan⁡2A1−tan⁡2A= \frac{(1 + \tan A)^2 + (1 - \tan A)^2}{1 - \tan^2A} = \frac{2 + 2\tan^2A}{1 - \tan^2A}

    =2(sin⁡2A+cos⁡2A)cos⁡2A−sin⁡2A=2cos⁡2A=2sec⁡2A== \frac{2(\sin^2A + \cos^2A)}{\cos^2A - \sin^2A} = \frac{2}{\cos 2A} = 2\sec 2A = R.H.S.

  42. We have to prove that 3cosec⁡20∘−sec⁡20∘=4\sqrt{3}\cosec 20^\circ - \sec 20^\circ = 4

    L.H.S. =3sin⁡20∘−1cos⁡20∘= \frac{\sqrt{3}}{\sin20^\circ} - \frac{1}{\cos20^\circ}

    =4(32)cos⁡20∘−12sin⁡20∘2sin⁡20∘cos⁡20∘= \frac{4(\frac{\sqrt{3}}{2})\cos20^\circ - \frac{1}{2}\sin20^\circ}{2\sin20^\circ\cos^20\circ}

    =4(sin⁡(50∘−20∘))sin⁡40∘=4== \frac{4(\sin(50^\circ - 20^\circ))}{\sin40^\circ} = 4 = R.H.S.

  43. We have to prove that tan⁡A+2tan⁡2A+4tan⁡4A+8cot⁡8A=cot⁡A\tan A + 2\tan 2A + 4\tan 4A + 8\cot 8A = \cot A

    tan⁡A−cot⁡A=sin⁡2A−cos⁡2Asin⁡Acos⁡A=−2cos⁡2Asin⁡2A=−2cot⁡2A\tan A - \cot A = \frac{\sin^2A - \cos^2A}{\sin A\cos A} = -\frac{2\cos 2A}{\sin 2A} = -2\cot 2A

    Similarly, 2tan⁡2A−2cot⁡2A=−4cot⁡4A2\tan 2A - 2\cot 2A = -4\cot 4A

    and 4tan⁡4A−4cot⁡4A=−8cot⁡8A4\tan 4A - 4\cot 4A = -8\cot 8A

    Thus, tan⁡A+2tan⁡2A+4tan⁡4A+8cot⁡8A=cot⁡A\tan A + 2\tan 2A + 4\tan 4A + 8\cot 8A = \cot A

  44. We have to prove that cos⁡2A+cos⁡2(2π3−A)+cos⁡2(2π3+A)=32\cos^2A + \cos^2\left(\frac{2\pi}{3} - A\right) + \cos^2\left(\frac{2\pi}{3} + A\right) = \frac{3}{2}

    ⇒2cos⁡2A+2cos⁡2(2π3−A)+2cos⁡2(2π3+A)=3\Rightarrow 2\cos^2A + 2\cos^2\left(\frac{2\pi}{3} - A\right) + 2\cos^2\left(\frac{2\pi}{3} + A\right) = 3

    L.H.S. =cos⁡2A+1+cos⁡(4π3−2A)+1+cos⁡(4π3+2A)+1= \cos 2A + 1 + \cos\left(\frac{4\pi}{3} - 2A\right) + 1 + \cos\left(\frac{4\pi}{3} + 2A\right) + 1

    =3+cos⁡2A+2cos⁡(4π3)cos⁡2A=3== 3 + \cos2A + 2\cos\left(\frac{4\pi}{3}\right)\cos2A = 3 = R.H.S.

  45. 2sin⁡2A+4cos⁡(A+B)sin⁡Asin⁡B+cos⁡2(A+B)2\sin^2A + 4\cos (A + B)\sin A\sin B + \cos2(A + B)

    =2sin⁡2A+2cos⁡(A+B)2sin⁡Asin⁡B+cos⁡2(A+B)= 2\sin^2A + 2\cos(A + B)2\sin A\sin B + \cos2(A + B)

    =2sin⁡2A+2cos⁡(A+B)[cos⁡(A−B)−cos⁡(A+B)]+cos⁡2(A+B)= 2\sin^2A + 2\cos(A + B)[\cos(A - B) - \cos(A + B)] + \cos2(A + B)

    =2sin⁡2A+2cos⁡(A+B)cos⁡(A−B)−2cos⁡2(A+B)+cos⁡2(A+B)= 2\sin^2A + 2\cos(A + B)\cos(A - B) - 2\cos^2(A + B) + \cos2(A + B)

    =2sin⁡2A+2(cos⁡2A−sin⁡2B)−2cos⁡2(A+B)+2cos⁡2(A+B)−1= 2\sin^2A + 2(\cos^2A - \sin^2B) - 2\cos^2(A + B) + 2\cos^2(A + B) - 1

    =2(sin⁡2A+cos⁡2A)−2sin⁡2B−1=1−2sin⁡2B= 2(\sin^2A + \cos^2A) -2\sin^2B - 1 = 1 -2\sin^2B which is independent of AA

  46. Given, cos⁡A=12(a+1a)\cos A = \frac{1}{2}\left(a + \frac{1}{a}\right)

    cos⁡2A=2cos⁡2A−1=2.14(a+1a)2−1\cos 2A = 2\cos^2A - 1 = 2.\frac{1}{4}\left(a + \frac{1}{a}\right)^2 - 1

    =12(a2+1a2)= \frac{1}{2}\left(a^2 + \frac{1}{a^2}\right)

  47. We have to prove that cos⁡2A+sin⁡2Acos⁡2B=cos⁡2B+sin⁡2Bcos⁡2A\cos^2A + \sin^2A\cos 2B = \cos^2B + \sin^2B\cos 2A

    ⇒cos⁡2A−cos⁡2B=sin⁡2Bcos⁡2A−sin⁡2Acos⁡2B\Rightarrow \cos^2A - \cos^2B = \sin^2B\cos2A - \sin^2A\cos2B

    R.H.S. =sin⁡2Bcos⁡2A−sin⁡2Acos⁡2B= \sin^2B\cos2A - \sin^2A\cos2B

    =sin⁡2B(cos⁡2A−sin⁡2A)−sin⁡2A(cos⁡2B−sin⁡2B)= \sin^2B(\cos^2A - \sin^2A) - \sin^2A(\cos^2B - \sin^2B)

    =cos⁡2Asin⁡2B−sin⁡2Acos⁡2B=cos⁡2A(1−cos⁡2B)−(1−cos⁡2A)cos⁡2B= \cos^2A\sin^2B - \sin^2A\cos^2B = \cos^2A(1 - \cos^2B) - (1 - \cos^2A)\cos^2B

    =cos⁡2A−cos⁡2B== \cos^2A - \cos^2B = R.H.S.

  48. We have to prove that 1+tan⁡Atan⁡2A=sec⁡2A1 + \tan A\tan 2A = \sec 2A

    L.H.S. =1+tan⁡Atan⁡2A=1+tan⁡A.2tan⁡A1−tan⁡2A= 1 + \tan A\tan 2A = 1 + \tan A.\frac{2\tan A}{1 - \tan^2A}

    =1+tan⁡2A1−tan⁡2A=cos⁡2A+sin⁡2Acos⁡2A−sin⁡2A= \frac{1 + \tan^2A}{1 - \tan^2A} = \frac{\cos^2A + \sin^2A}{\cos^2A - \sin^2A}

    =1cos⁡2A=sec⁡2A== \frac{1}{\cos 2A} = \sec 2A = R.H.S.

  49. We have to prove that 1+sin⁡2A1−sin⁡2A=(1+tan⁡A1−tan⁡A)2\frac{1 + \sin 2A}{1 - \sin 2A} = \left(\frac{1 + \tan A}{1 - \tan A}\right)^2

    L.H.S. =1+sin⁡2A1−sin⁡2A=sin⁡2A+cos⁡2A+2sin⁡Acos⁡Asin⁡2A+cos⁡2A−2sin⁡Acos⁡A= \frac{1 + \sin 2A}{1 - \sin 2A} = \frac{\sin^2A + \cos^2A + 2\sin A\cos A}{\sin^2A + \cos^2A - 2\sin A\cos A}

    =(sin⁡A+cos⁡Asin⁡A−cos⁡A)2= \left(\frac{\sin A + \cos A}{\sin A - \cos A}\right)^2

    Dividing numerator and denominator by cos⁡2A,\cos^2A, we get

    =(1+tan⁡A1−tan⁡A)2== \left(\frac{1 + \tan A}{1 - \tan A}\right)^2 = R.H.S.

  50. We have to prove that 1sin⁡10∘−3cos⁡10∘=4\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ} = 4

    L.H.S. =1sin⁡10∘−3cos⁡10∘= \frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}