Question

plz solve this problem ​

Answer 1

✨Given:✨

$\dfrac{1-\cos\theta}{\sin\theta}=\dfrac{\sin\theta}{1+\cos\theta}$

✨To Calculate:✨

To Prove LHS = RHS

✨Answer:✨

First, we will solve LHS.

LHS:

We can write sin $\theta$ as $\sqrt{1-\cos^2\theta}$.

$\\\dfrac{1-\cos\theta}{\sqrt{1-\cos^2\theta}}$

Squaring LHS, we get :

$\dfrac{(1-\cos\theta)^2}{1-\cos^2\theta}$

We know that,

$a^2-b^2=(a+b)(a-b)$

$\dfrac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)}\\\\\dfrac{1-\cos\theta}{1+\cos\theta}$

Multiplying numerator and denominator by (1+cos$\theta$), we get:

$\dfrac{1-\cos\theta}{1+\cos\theta}\times\dfrac{1+\cos\theta}{1+\cos\theta}\\\\\dfrac{1-\cos^2\theta}{(1+\cos\theta)^2}$

We know that,

(1-$\cos^2\theta$)=$\sin^2\theta$

$\dfrac{\sin^2\theta}{(1-\cos\theta)^2}\\\\\left(\dfrac{\sin\theta}{1+\cos\theta}\right)^2$

Square rooting LHS , we get:

$\dfrac{\sin\theta}{1+\cos\theta}$

LHS = RHS

Hence Proved.

⭐Other Trigonometric

Identities:⭐

$\\\\\sf1)\:{\sin}^{2}\theta+{\cos}^{2}\theta=1\\ \sf 2) \: {\sec}^{2} \theta - {\tan}^{2} \theta = 1\\ \sf 3) \: {\csc}^{2} \theta - {\cot}^{2} \theta = 1$

⭐Trigonometric Ratios:⭐

$\\\\ \tt 1) \: \sin\theta = \frac{1}{ \csc \theta} \\ \tt 2) \: \cos \: \theta \: \: = \: \frac{1}{\sec\theta} \\ \tt 3) \: \tan \: \theta \: = \: \frac{1}{\cot \: \theta} \\ \tt 4) \: \cot \: \theta \: = \: \frac{1}{\tan \: \theta} \\ \tt 5) \: \sec \: \theta \: = \: \frac{1}{\cos \: \theta} \\ \tt 6) \:\csc\:\theta= \frac{1}{\sin \: \theta}$