Question

PROVE THAT
:
1+secA/secA=sin^2A/1-cosA​

Answer 1

✨Given:✨

$\dfrac{1+\sec a}{\sec a}$ = $\dfrac{\sin^2 a}{1-\cos a}$

✨Solution:✨

First we will solve LHS.

LHS

$\dfrac{1+\sec a}{\sec a}$

$\dfrac{1}{\sec a}+\dfrac{\sec a}{\sec a}$

We know that,

$\dfrac{1}{\sec a}=\cos a$

$\cos a +1$

Now, we will solve RHS.

RHS

$\dfrac{\sin^2a}{1-\cos a}$

We know that,

$\sin^2a=1-\cos^2a$

$\dfrac{1-\cos^2a}{1-\cos a}$

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

$\dfrac{\cancel{(1-\cos a)}(1+\cos a)}{\cancel{(1-\cos a)}}$

$(\cos a+1)$

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:⭐

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