8. Compound Angles#

Algebraic sum of two or more angles is called a compound angle. If A,B,CA, B, C are any angle then A+B,A−B,A−B+C,A+B+C,A−B−C,A+B−CA + B, A - B, A - B + C, A + B + C, A - B - C, A + B -C etc. are all compound angles.

8.1. The Addition Formula#

sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin(A + B) = \sin A\cos B + \sin B\cos A

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A\cos B - \sin A\sin B

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

compound angle addition

Consider the diagram above. PMPM and PNPN are perpendicualr to OQOQ and ONON. RNRN is parallel to OQOQ and NQNQ is perpendicular to OQOQ. The left diagram represents the case when sum of angles is an acute angle while the right diagram represents the case when sum of angles is an obtuse angle.

∠RPN=90∘−∠PNR=∠RNO=∠NOQ=∠A\angle RPN = 90^{\circ} - \angle PNR = \angle RNO = \angle NOQ = \angle A

Now we can write, sin⁡(A+B)=sin⁡QOP=MPOP=MR+RPOP=QNOP+RPOP\sin(A + B) = \sin QOP = \frac{MP}{OP} = \frac{MR + RP}{OP} = \frac{QN}{OP} + \frac{RP}{OP}

=QNONONOP+RPNPNPOP=sin⁡Acos⁡B+cos⁡Asin⁡B=\frac{QN}{ON}\frac{ON}{OP} + \frac{RP}{NP}\frac{NP}{OP} = \sin A\cos B + \cos A\sin B

Also, cos⁡(A+B)=cos⁡QOP=OMOP=OQ−MQOP=OQONONOP−RNNPNPOP\cos(A + B) = \cos QOP = \frac{OM}{OP} = \frac{OQ - MQ}{OP} = \frac{OQ}{ON}\frac{ON}{OP} - \frac{RN}{NP}\frac{NP}{OP}

=cos⁡Acos⁡B−sin⁡Asin⁡B= \cos A\cos B - \sin A\sin B

These two results lead to tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}

We have shown that addition formula is true when angles involved are acute angles. The same proof can be applied to prove the results for all values of AA and BB.

Consider A′=90∘+A∴‘sin⁡A′=cos⁡AA' = 90^{\circ} + A \therefore `\sin A' = \cos A and cos⁡A′=sin⁡A\cos A' = \sin A

sin⁡(A′+B)=cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B=sin⁡A′cos⁡B+cos⁡A′sin⁡B\sin(A' + B) = \cos (A + B) = \cos A\cos B - \sin A\sin B = \sin A'\cos B + \cos A'\sin B

Similarly cos⁡(A′+B)=−sin⁡(A+B)=−sin⁡Acos⁡B−sin⁡Bcos⁡A=cos⁡A′cos⁡B−sin⁡A′sin⁡B\cos(A' + B) = -\sin(A + B) = -\sin A\cos B - \sin B\cos A = \cos A'\cos B - \sin A'\sin B

We can prove it again for B′=90∘+BB' = 90^{\circ} + B and so on by increasing the values of AA and BB. Then we can again increase values by 90∘90^{\circ} and proceeding this way we see that the formula holds true for all values of AA and BB.

8.2. The Subtraction Formula#

sin⁡(A−B)=sin⁡Acos⁡B−sin⁡Bcos⁡A\sin(A - B) = \sin A\cos B - \sin B\cos A

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A\cos B + \sin A\sin B

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

compound angle subtraction

Conside the diagram above. The angle MOPMOP is A−B.A - B. We take a point P,P, and draw PMPM and PNPN perpendicular to OMOM and ONON respectively. From NN we draw NQNQ and NRNR perpendicular to OQOQ and MPMP respectively.

∠RPN=90∘−∠PNR=∠QON=A\angle RPN = 90^{\circ} - \angle PNR = \angle QON = A

Thus, we can write sin⁡(A−B)=sin⁡MOP=MPOP=MR−PROP=QNONONOP−PRPNPNOP\sin(A - B) = \sin MOP = \frac{MP}{OP} = \frac{MR - PR}{OP} = \frac{QN}{ON}\frac{ON}{OP} - \frac{PR}{PN}\frac{PN}{OP}

Thus, sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A - B) = \sin A\cos B - \cos A\sin B

Also, cos⁡(A−B)=OMOP=OQ+QMOP=OQONONOP+RNNPNPOP\cos(A - B) = \frac{OM}{OP} = \frac{OQ + QM}{OP} = \frac{OQ}{ON}\frac{ON}{OP} + \frac{RN}{NP}\frac{NP}{OP}

=cos⁡Acos⁡B+sin⁡Asin⁡B= \cos A\cos B + \sin A\sin B

We have shown that subtraction formula is true when angles involved are acute angles. The same proof can be applied to prove the results for all values of AA and BB.

From the results obtained we find upon division that tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}

8.3. Important Deductions#

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

    Proof: L.H.S. =(sin⁡Acos⁡B+sin⁡Bcos⁡A)(sin⁡Acos⁡B−sin⁡Bcos⁡A)= (\sin A\cos B + \sin B\cos A)(\sin A\cos B - \sin B\cos A)

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

    =sin⁡2A−sin⁡2Asin⁡2B−sin⁡2B+sin⁡2Bsin⁡2A= \sin^2A - \sin^2A\sin^2B - \sin^2B + \sin^2B\sin^2A

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

    =cos⁡2B−cos⁡2A= \cos^2B - \cos^2A

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

    Proof: L.H.S. =(cos⁡Acos⁡B−sin⁡Asin⁡B)(cos⁡Acos⁡B+sin⁡Asin⁡B)=(\cos A\cos B - \sin A\sin B)(\cos A\cos B + \sin A\sin B)

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

    =cos⁡2A−cos⁡2Asin⁡2B−sin⁡2B+cos⁡2Asin⁡2B=cos⁡2A−sin⁡2B=cos⁡2B−sin⁡2A=\cos^2A - \cos^2A\sin^2B - \sin^2B + \cos^2A\sin^2B = \cos^2A - \sin^2B = \cos^2B - \sin^2A

  3. cot⁡(A+B)=cot⁡Acot⁡B−1cot⁡B+cot⁡A\cot(A + B) = \frac{\cot A\cot B - 1}{\cot B + \cot A}

    Proof: L.H.S. =cot⁡(A+B)=cos⁡(A+B)sin⁡(A+B)= \cot(A + B) = \frac{\cos(A + B)}{\sin(A + B)}

    =cos⁡Acos⁡B−sin⁡Asin⁡Bsin⁡Acos⁡B+cos⁡Asin⁡B= \frac{\cos A\cos B - \sin A\sin B}{\sin A\cos B + \cos A\sin B}

    Dividing numberator and denominator by sin⁡Asin⁡B\sin A\sin B

    =cot⁡Acot⁡B−1cot⁡B+cot⁡A= \frac{\cot A\cot B - 1}{\cot B + \cot A}

  4. cot⁡(A−B)=cot⁡Acot⁡B+1cot⁡B−cot⁡A\cot(A - B) = \frac{\cot A\cot B + 1}{\cot B - \cot A}

    Proof: L.H.S. =cot⁡(A−B)=cos⁡(A−B)sin⁡(A−B)= \cot(A - B) = \frac{\cos(A - B)}{\sin(A - B)}

    =cos⁡Acos⁡B+sin⁡Asin⁡Bsin⁡Acos⁡B−cos⁡Asin⁡B= \frac{\cos A\cos B + \sin A\sin B}{\sin A\cos B - \cos A\sin B}

    Dividing numberator and denominator by sin⁡Asin⁡B\sin A\sin B

    =cot⁡Acot⁡B+1cot⁡B−cot⁡A= \frac{\cot A\cot B + 1}{\cot B - \cot A}

  5. tan⁡(A+B+C)=tan⁡A+tan⁡B+tan⁡C−tan⁡Atan⁡Btan⁡C1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Ctan⁡A\tan(A + B + C) = \frac{\tan A + \tan B + \tan C - \tan A\tan B\tan C}{1 - \tan A\tan B - \tan B\tan C - \tan C\tan A}

    Proof: L.H.S. =tan⁡[(A+B)+C]=tan⁡(A+B)+tan⁡C1−tan⁡(A+B)tan⁡C= \tan[(A + B) + C] = \frac{\tan(A + B) + \tan C}{1 - \tan(A + B)\tan C}

    =tan⁡A+tan⁡B1−tan⁡Atan⁡B+tan⁡C1−tan⁡A+tan⁡B1−tan⁡Atan⁡Btan⁡C= \frac{\frac{\tan A + \tan B}{1 - \tan A\tan B} + \tan C}{1 - \frac{\tan A + \tan B}{1 - \tan A\tan B}\tan C}

    =tan⁡A+tan⁡B+tan⁡C−tan⁡Atan⁡Btan⁡C1−tan⁡Atan⁡B1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Ctan⁡A1−tan⁡Atan⁡B= \frac{\frac{\tan A + \tan B + \tan C - \tan A\tan B\tan C}{1 - \tan A\tan B}}{\frac{1 - \tan A\tan B - \tan B\tan C - \tan C\tan A}{1 - \tan A\tan B}}

    =tan⁡A+tan⁡B+tan⁡C−tan⁡Atan⁡Btan⁡C1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Ctan⁡A= \frac{\tan A + \tan B + \tan C - \tan A\tan B\tan C}{1 - \tan A\tan B - \tan B\tan C - \tan C\tan A}

8.4. To express acos⁡θ+bsin⁡θa\cos\theta + b\sin\theta in the form of kcos⁡ϕk\cos\phi or ksin⁡ϕk\sin\phi#

acos⁡θ+bsin⁡θ=a2+b2(aa2+b2cos⁡θ+ba2+b2sin⁡θ)a\cos\theta + b\sin\theta = \sqrt{a^2 + b^2}\left(\frac{a}{\sqrt{a^2 + b^2}}\cos\theta + \frac{b}{\sqrt{a^2 + b^2}}\sin\theta\right)

Let cos⁡α=aa2+b2\cos\alpha = \frac{a}{\sqrt{a^2 + b^2}} then sin⁡α=ba2+b2\sin\alpha = \frac{b}{\sqrt{a^2 + b^2}}

Thus, acos⁡θ+bsin⁡θ=a2+b2(cos⁡αcos⁡θ+sin⁡αsin⁡θ)a\cos\theta + b\sin\theta = \sqrt{a^2 + b^2}(\cos\alpha\cos\theta + \sin\alpha\sin\theta)

=a2+b2cos⁡(θ−α)=kcos⁡ϕ= \sqrt{a^2 + b^2}\cos(\theta - \alpha) = k\cos\phi where k=a2+b2k = \sqrt{a^2 + b^2} and ϕ=θ−α\phi = \theta - \alpha

Alternatively, if aa2+b2=sin⁡α\frac{a}{\sqrt{a^2 + b^2}} = \sin\alpha then ba2+b2=cos⁡α\frac{b}{\sqrt{a^2 + b^2}} = \cos\alpha

Thus, acos⁡θ+bsin⁡θ=a2+b2(sin⁡αcos⁡θ+cos⁡α+sin⁡θ)a\cos\theta + b\sin\theta = \sqrt{a^2 + b^2}(\sin\alpha\cos\theta + \cos\alpha + \sin\theta)

=a2+b2sin⁡(θ+α)=ksin⁡ϕ= \sqrt{a^2 + b^2}\sin(\theta + \alpha) = k\sin\phi where k=a2+b2k = \sqrt{a^2+b^2} and ϕ=θ+α\phi = \theta + \alpha

8.5. Problems#

  1. If sin⁡α=35\sin\alpha = \frac{3}{5} and cos⁡β=941,\cos\beta = \frac{9}{41}, find the values of sin⁡(α−β)\sin(\alpha - \beta) and cos⁡(α+β).\cos(\alpha + \beta).

  2. If sin⁡α=4553\sin\alpha = \frac{45}{53} and sin⁡β=3365,\sin\beta = \frac{33}{65}, find the values of sin⁡(α−β)\sin(\alpha - \beta) and sin⁡(α+β).\sin(\alpha + \beta).

  3. If sin⁡α=1517\sin\alpha = \frac{15}{17} and cos⁡β=1213,\cos\beta = \frac{12}{13}, find the values of sin⁡(α+β),cos⁡(α−β)\sin(\alpha + \beta), \cos(\alpha - \beta) and tan⁡(α+beta).\tan(\alpha + beta).

Prove the following:

  1. cos⁡(45∘−A)cos⁡(45∘−B)−sin⁡(45∘−A)sin⁡(45∘−B)=sin⁡(A+B)\cos(45^{\circ} - A)\cos(45^{\circ} - B) - \sin(45^{\circ} - A)\sin(45^{\circ} - B) = \sin(A + B)

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

  3. sin⁡(A−B)cos⁡Acos⁡B+sin⁡(B−C)cos⁡Bcos⁡C+sin⁡(C−A)cos⁡Ccos⁡A=0\frac{\sin(A - B)}{\cos A\cos B} + \frac{\sin(B - C)}{\cos B\cos C} + \frac{\sin(C - A)}{\cos C\cos A} = 0

  4. sin⁡105∘+cos⁡105∘=cos⁡45∘\sin 105^\circ + \cos 105^\circ = \cos 45^\circ

  5. sin⁡75∘−sin⁡15∘=cos⁡105∘+cos⁡15∘\sin 75^\circ - \sin 15^\circ = \cos 105^\circ + \cos 15^\circ

  6. cos⁡αcos⁡(γ−α)−sin⁡αsin⁡(γ−α)=cos⁡γ\cos\alpha\cos(\gamma - \alpha) - \sin\alpha\sin(\gamma - \alpha) = \cos\gamma

  7. cos⁡(α+β)cos⁡γ−cos⁡(β+γ)cos⁡α=sin⁡βsin⁡(γ−α)\cos(\alpha + \beta)\cos\gamma - \cos(\beta + \gamma)\cos\alpha = \sin\beta\sin(\gamma - \alpha)

  8. sin⁡(n+1)Asin⁡(n−1)A+cos⁡(n+1)Acos⁡(n−1)A=cos⁡2A\sin(n + 1)A\sin(n - 1)A + \cos(n + 1)A\cos(n - 1)A = \cos 2A

  9. sin⁡(n+1)Asin⁡(n+2)A+cos⁡(n+1)Acos⁡(n+2)A=cos⁡A\sin(n + 1)A\sin(n + 2)A + \cos(n + 1)A\cos(n + 2)A = \cos A

  10. Find the value of cos⁡15∘\cos 15^\circ and sin⁡105∘\sin 105^\circ

  11. Find the value of tan⁡105∘\tan 105^\circ

  12. Find the value of tan⁡495∘cot⁡855∘\frac{\tan 495^\circ}{\cot 855^\circ}

  13. Evaluate sin⁡(nπ+(−1)nπ4),\sin\left(n\pi + (-1)^n \frac{\pi}{4}\right), where nn is an integer.

Prove the following:

  1. sin⁡15∘=3−122\sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}}

  2. cos⁡75∘=3−122\cos 75^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}}

  3. tan⁡75∘=2+3\tan 75^\circ = 2 + \sqrt{3}

  4. tan⁡15∘=2−3\tan 15^\circ = 2 - \sqrt{3}

Find the value of following:

  1. cos⁡1395∘\cos 1395^\circ

  2. tan⁡(−330∘)\tan(-330^\circ)

  3. sin⁡300∘cosec⁡1050∘−tan⁡(−120∘)\sin 300^\circ \cosec 1050^\circ - \tan(-120^\circ)

  4. tan⁡(11π12)\tan\left(\frac{11\pi}{12}\right)

  5. tan⁡((−1)nπ4)\tan \left((-1)^n\frac{\pi}{4}\right)

Prove the following:

  1. cos⁡18∘−sin⁡18∘=2sin⁡27∘\cos 18^\circ - \sin 18^\circ = \sqrt{2}\sin 27^\circ

  2. tan⁡70∘=2tan⁡50∘+tan⁡20∘\tan 70^\circ = 2\tan 50^\circ + \tan 20^\circ

  3. cot⁡(π4+x)cot⁡(π4−x)=1\cot\left(\frac{\pi}{4} + x\right)\cot\left(\frac{\pi}{4} - x\right) = 1

  4. cos⁡(m+n)θ.cos⁡(m−n)θ−sin⁡(m+n)θsin⁡(m−n)θ=cos⁡2mθ\cos(m + n)\theta.\cos(m - n)\theta - \sin(m + n)\theta\sin(m - n)\theta = \cos 2m\theta

  5. tan⁡(θ+ϕ)+tan⁡(θ−ϕ)1−tan⁡(θ+ϕ)tan⁡(θ−ϕ)=tan⁡2θ\frac{\tan(\theta + \phi) + \tan(\theta - \phi)}{1 - \tan(\theta + \phi)\tan(\theta - \phi)} = \tan 2\theta

  6. cos⁡9∘+sin⁡9∘=2sin⁡54∘\cos 9^\circ + \sin 9^\circ = \sqrt{2}\sin 54^\circ

  7. cos⁡20∘−sin⁡20∘cos⁡20∘+sin⁡20∘=tan⁡25∘\frac{\cos 20^\circ - \sin 20^\circ}{\cos 20^\circ + \sin 20^\circ} = \tan 25^\circ

  8. tan⁡A+tan⁡Btan⁡A−tan⁡B=sin⁡(A+B)sin⁡(A−B)\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\sin(A + B)}{\sin(A - B)}

  9. 1tan⁡3A−tan⁡A−1cot⁡3A−cot⁡A=cot⁡2A\frac{1}{\tan 3A - \tan A} - \frac{1}{\cot 3A - \cot A} = \cot 2A

  10. 1tan⁡3A+tan⁡A−1cot⁡3A−cot⁡A=cot⁡4A\frac{1}{\tan 3A + \tan A} - \frac{1}{\cot 3A - \cot A} = \cot 4A

  11. sin⁡3αsin⁡α+cos⁡3αcosα=4cos⁡2α\frac{\sin 3\alpha}{\sin\alpha} + \frac{\cos 3\alpha}{cos\alpha} = 4\cos 2\alpha

  12. tan⁡(π4+A)−tan⁡(π4−A)tan⁡(π4+A)+tan⁡(π4−A)=sin⁡2A\frac{\tan\left(\frac{\pi}{4} + A \right) - \tan\left(\frac{\pi}{4} - A\right)}{\tan\left(\frac{\pi}{4} + A\right) + \tan\left(\frac{\pi}{4} - A\right)} = \sin 2A

  13. tan⁡40∘+2tan⁡10∘=tan⁡50∘\tan 40^\circ + 2 \tan 10^\circ = \tan 50^\circ

  14. tan⁡(α+β)tan⁡(α−β)=sin⁡2α−sin⁡2βcos⁡2α−sin⁡2β\tan(\alpha + \beta)\tan(\alpha - \beta) = \frac{\sin^2\alpha - \sin^2\beta}{\cos^2\alpha - \sin^2\beta}

  15. tan⁡2α−tan⁡2β=sin⁡(α+β)sin⁡(α−β)cos⁡2αcos⁡2β\tan^2\alpha -\tan^2\beta = \frac{\sin(\alpha + \beta)\sin(\alpha - \beta)}{\cos^2\alpha\cos^2\beta}

  16. tan⁡[(2n+1)π+θ]+tan⁡[(2n+1)π−θ]=0\tan[(2n + 1)\pi + \theta] + \tan[(2n + 1)\pi - \theta] = 0

  17. tan⁡(π4+θ)tan⁡(3π4+θ)+1=0\tan\left(\frac{\pi}{4} + \theta\right)\tan\left(\frac{3\pi}{4} + \theta\right) + 1 = 0

  18. If tan⁡α=p\tan\alpha = p and tan⁡β=q\tan\beta = q prove that cos⁡(α+β)=1−pq(1+p2)(1+q2)\cos(\alpha + \beta) = \frac{1 - pq}{\sqrt{(1 + p^2)(1 + q^2)}}

  19. if tan⁡β=2sin⁡αsin⁡γsin⁡(α+γ),\tan \beta = \frac{2\sin\alpha\sin\gamma}{\sin(\alpha + \gamma)}, show that cot⁡α,cot⁡β,cot⁡γ\cot\alpha, \cot\beta, \cot\gamma are in A.P.

  20. Eliminate θ\theta if tan⁡(θ−α)=a\tan(\theta - \alpha) = a and tan⁡(θ+α)=b\tan(\theta + \alpha) = b

  21. Eliminate α\alpha and β\beta if tan⁡α+tan⁡β=b,cot⁡α+cot⁡β=a\tan\alpha + \tan\beta = b, \cot\alpha + \cot\beta = a and α+β=γ\alpha + \beta = \gamma

  22. If A+B=45∘,A + B = 45^\circ, show that (1+tan⁡A)(1+tan⁡B)=2(1 + \tan A)(1 + \tan B) = 2

  23. If sin⁡αsin⁡β−cos⁡αcos⁡β+1=0,\sin\alpha\sin\beta - \cos\alpha\cos\beta + 1 = 0, prove that 1+cot⁡αtan⁡β=01 + \cot\alpha\tan\beta = 0

  24. If tan⁡β=nsin⁡αcos⁡α1−nsin⁡2α,\tan\beta = \frac{n\sin\alpha\cos\alpha}{1 - n\sin^2\alpha}, prove that tan⁡(α−β)=(1−n)α\tan(\alpha - \beta) = (1 - n)\alpha

  25. If cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)=−32,\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = -\frac{3}{2}, prove that cos⁡α+cos⁡β+cos⁡γ=sin⁡α+sin⁡β+sin⁡γ=0\cos\alpha + \cos\beta + \cos\gamma = \sin\alpha + \sin\beta + \sin\gamma = 0

  26. If tan⁡α=mm+1,tan⁡β=12m+1,\tan\alpha = \frac{m}{m + 1}, \tan\beta = \frac{1}{2m + 1}, prove that α+β=π4\alpha + \beta = \frac{\pi}{4}

  27. If A+B=45∘,A + B = 45^\circ, show that (cot⁡A−1)(cot⁡B−1)=2(\cot A - 1)(\cot B - 1) = 2

  28. If tan⁡α−tan⁡β=x\tan\alpha - \tan\beta = x and cot⁡β−cot⁡α=y,\cot\beta - \cot\alpha = y, prove that cot⁡(α−β)=x+yxy\cot(\alpha - \beta) = \frac{x + y}{xy}

  29. If a right angle be divided into three pats α,β\alpha, \beta and γ,\gamma, prove that cot⁡α=tan⁡β+tan⁡γ1−tan⁡βtan⁡γ\cot\alpha = \frac{\tan\beta + \tan\gamma}{1 - \tan\beta\tan\gamma}

  30. If 2tan⁡β+cot⁡β=tan⁡α,2\tan\beta + \cot \beta = \tan\alpha, show that cot⁡β=2tan⁡(α−β)\cot \beta = 2\tan(\alpha - \beta)

  31. If in any △ABC,C=90∘,\triangle ABC, C = 90^\circ, prove that cosec⁡(A−B)=a2+b2a2−b2\cosec(A - B) = \frac{a^2 + b^2}{a^2 - b^2} and sec⁡(A−B)=c22ab\sec(A - B) = \frac{c^2}{2ab}

  32. If cot⁡A=ac,cot⁡B=ca,tan⁡C=ca3\cot A = \sqrt{ac}, \cot B = \sqrt{\frac{c}{a}}, \tan C = \sqrt{\frac{c}{a^3}} and c=a2+a+1,c = a^2 + a + 1, prove that A=B+CA = B + C

  33. If tan⁡(A−B)tan⁡A+sin⁡2Csin⁡2A=1,\frac{\tan(A - B)}{\tan A} + \frac{\sin^2C}{\sin^2A} = 1, prove that tan⁡Atan⁡B=tan⁡2C\tan A\tan B = \tan^2 C

  34. If sin⁡αsin⁡β−cos⁡αcos⁡β=1\sin\alpha\sin\beta - \cos\alpha\cos\beta = 1 show that tan⁡α+tan⁡β=0\tan\alpha + \tan\beta = 0

  35. If sin⁡θ=3sin⁡(θ+2α),\sin\theta = 3\sin(\theta + 2\alpha), prove that tan⁡(θ+α),\tan(\theta + \alpha), prove that tan⁡(θ+α)+2tan⁡α=0\tan(\theta + \alpha) + 2\tan\alpha = 0

  36. If 3tan⁡θtan⁡ϕ=1,3\tan\theta\tan\phi = 1, prove that 2cos⁡(θ+ϕ)=cos⁡(θ−α)2\cos(\theta + \phi) = \cos(\theta - \alpha)

  37. Find the sign of the expression sin⁡θ+cos⁡θ\sin\theta + \cos\theta when θ=100∘\theta = 100^\circ

  38. Prove that the value of 5cos⁡θ+3cos⁡(θ+π3)+35\cos\theta + 3\cos\left(\theta + \frac{\pi}{3}\right) + 3 lies between −4-4 and 1010

  39. If mtan⁡(θ−30∘)=ntan⁡(θ+120∘),m\tan(\theta - 30^\circ) = n\tan(\theta + 120^\circ), show that cos⁡2θ=m+n2(m−n)\cos2\theta = \frac{m + n}{2(m - n)}

  40. if α+β=θ\alpha + \beta = \theta and tan⁡α:tan⁡β=x:y,\tan\alpha:\tan\beta = x:y, prove that sin⁡(α−β)=x−yx+ysin⁡θ\sin(\alpha - \beta) = \frac{x - y}{x + y}\sin\theta

  41. Find the maximum and minimum value of 7cos⁡θ+24sin⁡θ7\cos\theta + 24\sin\theta

  42. Show that sin⁡100∘−sin⁡10∘\sin 100^\circ - \sin 10^\circ is positive.