Trignometry Exercises

  1. 29 May 2024

Trignometry Exercises for AL students

Proove below trignometric equations. (Basic Level)

1. (1−sin⁡θ)(1+sin⁡θ)=cos⁡θ\sqrt{(1-\sin \theta)(1+\sin \theta)} = \cos \theta

L.H.S:−L.H.S :-

=(1−sin⁡θ)(1+sin⁡θ)= \sqrt{(1-\sin \theta)(1+\sin \theta)}

=(12−sin⁡2θ)= \sqrt{(1^2-\sin^2 \theta)}

=cos⁡2θ= \sqrt{\cos^2 \theta}

=cos⁡θ= \cos \theta

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

2. (1−cos⁡θ)(1+cos⁡θ)=sin⁡θ\sqrt{(1-\cos \theta)(1+\cos \theta)} = \sin \theta

L.H.S:−L.H.S :-

=(1−cos⁡θ)(1+cos⁡θ)= \sqrt{(1-\cos \theta)(1+\cos \theta)}

=(12−cos⁡2θ)= \sqrt{(1^2-\cos^2 \theta)}

=sin⁡2θ= \sqrt{\sin^2 \theta}

=sin⁡θ= \sin \theta

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

3. 1cos⁡2θ+1cos⁡2θ=1sin⁡2θcos⁡2θ\frac{1}{\cos^2 \theta} + \frac{1}{\cos^2 \theta} = \frac{1}{\sin^2 \theta\cos^2 \theta}

L.H.S:−L.H.S :-

=1cos⁡2θ+1cos⁡2θ= \frac{1}{\cos^2 \theta} + \frac{1}{\cos^2 \theta}

=sin⁡2θ+cos⁡2θsin⁡2θcos⁡2θ= \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}

=1sin⁡2θcos⁡2θ= \frac{1}{\sin^2 \theta \cos^2 \theta}

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

4. cosec⁡2θ−1=cos⁡θcosec⁡θ\sqrt{\cosec^2 \theta - 1} = \cos \theta \cosec \theta

L.H.S:−L.H.S :-

=cosec⁡2θ−1= \sqrt{\cosec^2 \theta - 1}

=cot⁡2θ= \sqrt{\cot^2 \theta}

=cot⁡θ= \cot \theta

=cos⁡θcosec⁡θ= \cos \theta \cosec \theta

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

5. sec⁡2θ−1=sin⁡θsec⁡θ\sqrt{\sec^2 \theta - 1} = \sin \theta \sec \theta

L.H.S:−L.H.S :-

=sec⁡2θ−1= \sqrt{\sec^2 \theta - 1}

=tan⁡2θ= \sqrt{\tan^2 \theta}

=tan⁡θ= \tan \theta

=sin⁡θsec⁡θ= \sin \theta \sec \theta

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

6. sec⁡4θ−sec⁡2θ=tan⁡4θ+tan⁡2θ\sec^4 \theta - \sec^2 \theta = \tan^4 \theta + \tan^2 \theta

L.H.S:−L.H.S :-

=sec⁡4θ−sec⁡2θ= \sec^4 \theta - \sec^2 \theta

=1−cos⁡2θcos⁡4θcos⁡2θ= \frac{1 - \cos^2 \theta}{\cos^4 \theta \cos^2 \theta}

=sin⁡2θcos⁡4θcos⁡2θ= \frac{\sin^2 \theta}{\cos^4 \theta \cos^2 \theta}

=tan⁡2θsec⁡2θ= \tan^2 \theta \sec^2 \theta

=tan⁡2θ(1+tan⁡2θ)= \tan^2 \theta (1 + \tan^2 \theta)

=tan⁡4θ+tan⁡2θ= \tan^4 \theta + \tan^2 \theta

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

7. sin⁡2θ−sin⁡4θ=cos⁡2θ−cos⁡4θ\sin^2 \theta - \sin^4 \theta = \cos^2 \theta - \cos^4 \theta

L.H.S:−L.H.S :-

=sin⁡2θ−sin⁡4θ= \sin^2 \theta - \sin^4 \theta

=sin⁡2θ(1−sin⁡2θ)= \sin^2 \theta (1 - \sin^2 \theta)

=(1−cos⁡2θ)cos⁡2θ= (1 - \cos^2 \theta) \cos^2 \theta

=cos⁡2θ−cos⁡4θ= \cos^2 \theta - \cos^4 \theta

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

8. sec⁡4θ−sec⁡2θ=tan⁡4θ+tan⁡2θ\sec^4 \theta - \sec^2 \theta = \tan^4 \theta + \tan^2 \theta

L.H.S:−L.H.S :-

=sec⁡4θ−sec⁡2θ= \sec^4 \theta - \sec^2 \theta

=sec⁡2θ(sec⁡2θ−1)= \sec^2 \theta (\sec^2 \theta - 1)

=(1+tan⁡4θ)tan⁡2θ= (1 + \tan^4 \theta) \tan^2 \theta

=tan⁡4θ+tan⁡2θ= \tan^4 \theta + \tan^2 \theta

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

9. cot⁡4θ+cot⁡2θ=cosec⁡4θ−cosec⁡2θ\cot^4 \theta + \cot^2 \theta = \cosec^4 \theta - \cosec^2 \theta

L.H.S:−L.H.S :-

=cot⁡4θ+cot⁡2θ= \cot^4 \theta + \cot^2 \theta

=cot⁡2θ(cot⁡2θ+1)= \cot^2 \theta (\cot^2 \theta + 1)

=(cosec⁡2θ−1)cosec⁡2θ= (\cosec^2 \theta - 1) \cosec^2 \theta

=cosec⁡4θ−cosec⁡2θ= \cosec^4 \theta - \cosec^2 \theta

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

10. (1−sin⁡θ)(cosec⁡θ+1)=cos⁡θcot⁡θ(1 - \sin \theta)(\cosec \theta + 1) = \cos \theta \cot \theta

L.H.S:−L.H.S :-

=(1−sin⁡θ)(cosec⁡θ+1)= (1 - \sin \theta)(\cosec \theta + 1)

=(1−sin⁡θ)(1+sin⁡θsin⁡θ)= (1 - \sin \theta)(\frac{1 + \sin \theta}{\sin \theta})

=1−sin⁡2θsin⁡θ= \frac{1 - \sin^2 \theta}{\sin \theta}

=cos⁡2θsin⁡θ= \frac{\cos^2 \theta}{\sin \theta}

=cos⁡θcot⁡θ= \cos \theta \cot \theta

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

11. (1−cos⁡θ)(1+sec⁡θ)=sin⁡θtan⁡θ(1 - \cos \theta)(1 + \sec \theta) = \sin \theta \tan \theta

L.H.S:−L.H.S :-

=(1−cos⁡θ)(1+sec⁡θ)= (1 - \cos \theta)(1 + \sec \theta)

=(1−cos⁡θ)(cos⁡θ+1cos⁡θ)= (1 - \cos \theta)(\frac{\cos \theta + 1}{\cos \theta})

=1−cos⁡2θcos⁡θ= \frac{1 - \cos^2 \theta}{\cos \theta}

=sin⁡2θcos⁡θ= \frac{\sin^2 \theta}{\cos \theta}

=sin⁡θtan⁡θ= \sin \theta \tan \theta

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

12. sec⁡θ−tan⁡θ=1secθ+tanθ\sec \theta - \tan \theta = \frac{1}{sec \theta + tan \theta}

L.H.S:−L.H.S :-

=sec⁡θ−tan⁡θ= \sec \theta - \tan \theta

=sec⁡2θ−tan⁡2θsec⁡θ+tan⁡θ= \frac{\sec^2 \theta - \tan^2 \theta}{\sec \theta + \tan \theta}

=1secθ+tanθ= \frac{1}{sec \theta + tan \theta}

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

13. cosec⁡θ−cot⁡θ=1cosecθ−cotθ\cosec \theta - \cot \theta = \frac{1}{cosec \theta - cot \theta}

L.H.S:−L.H.S :-

=cosec⁡θ−cot⁡θ= \cosec \theta - \cot \theta

=cosec⁡2θ−cot⁡2θcosec⁡θ+cot⁡θ= \frac{\cosec^2 \theta - \cot^2 \theta}{\cosec \theta + \cot \theta}

=1cosecθ+cotθ= \frac{1}{cosec \theta + cot \theta}

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

Proove below trignometric equations considering A+B+C=πA + B + C = \pi for all the below questions. (Expert Level)

1. sin⁡A+sin⁡B+sin⁡C=4cos⁡A2cos⁡B2cos⁡C2\sin A + \sin B + \sin C = 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}

L.H.S:−L.H.S :-

=sin⁡A+sin⁡B+sin⁡C= \sin A + \sin B + \sin C

=2sin⁡(A+B2)cos⁡(A−B2)+2sin⁡C2cos⁡C2= 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2}) + 2\sin \frac{C}{2} \cos \frac{C}{2}

=2sin⁡(π−C2)cos⁡(A−B2)+2sin⁡C2cos⁡C2= 2\sin(\frac{\pi - C}{2})\cos(\frac{A-B}{2}) + 2\sin \frac{C}{2} \cos \frac{C}{2}

=2cos⁡C2[cos⁡(A−B2)+sin⁡C2]= 2\cos \frac{C}{2} [\cos(\frac{A-B}{2}) + \sin \frac{C}{2}]

=2cos⁡C2[cos⁡(A−B2)+sin⁡(π−A+B2)]= 2\cos \frac{C}{2} [\cos(\frac{A-B}{2}) + \sin(\frac{\pi - A + B}{2})]

=2cos⁡C2[cos⁡(A−B2)+cos⁡(A+B2)]= 2\cos \frac{C}{2} [\cos(\frac{A-B}{2}) + \cos(\frac{A + B}{2})]

=2cos⁡C2[2cos⁡A2cos⁡B2]= 2\cos \frac{C}{2} [2\cos \frac{A}{2} \cos \frac{B}{2}]

=4cos⁡A2cos⁡B2cos⁡C2= 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}

∴L.H.S=R.H.S\therefore L.H.S = R.H.S

2. sin⁡2A+sin⁡2B+sin⁡2C=2sin⁡Asin⁡Bcos⁡C\sin^2 A + \sin^2 B + \sin^2 C = 2\sin A \sin B \cos C

L.H.S:−L.H.S :-

=sin⁡2A+sin⁡2B+sin⁡2C= \sin^2 A + \sin^2 B + \sin^2 C

=sin⁡Asin⁡(π−(B+C))+(sin⁡B+sin⁡C)(sin⁡B−sin⁡C)= \sin A \sin(\pi - (B+C)) + (\sin B + \sin C)(\sin B - \sin C)

=sin⁡Asin⁡(B+C)+[2sin⁡(B+C2)cos⁡(B−C2)2sin⁡(B−C2)cos⁡(B+C2)]= \sin A \sin(B+C) + [2\sin(\frac{B+C}{2})\cos(\frac{B-C}{2})2\sin(\frac{B-C}{2})\cos(\frac{B+C}{2})]

=sin⁡Asin⁡(B+C)+[2sin⁡(B+C2)cos⁡(B+C2)2sin⁡(B−C2)cos⁡(B−C2)]= \sin A \sin(B+C) + [2\sin(\frac{B+C}{2})\cos(\frac{B+C}{2})2\sin(\frac{B-C}{2})\cos(\frac{B-C}{2})]

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

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

=sin⁡A[sin⁡(B+C)+sin⁡(B−C)]= \sin A [\sin(B+C) + \sin(B-C)]

=2sin⁡Asin⁡Bcos⁡C= 2\sin A \sin B \cos C

∴L.H.S=R.H.S\therefore L.H.S = R.H.S