Question

1 + cos theta + sin theta upon 1 + cos theta minus sin theta equal to 1 + sin theta upon cos theta

Answer 1

Answer:

Hence Proved

Step-by-step explanation:

To prove:

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

Proof: We start from left side,

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

Divide by $\cos\theta$ at top and denominator and we get

$\Rightarrow \dfrac{\sec\theta+1+\tan\theta}{\sec\theta+1-\tan\theta}$

$\Rightarrow \dfrac{\sec\theta+\tan\theta+\sec^2\theta-\tan^2\theta}{\sec\theta+1-\tan\theta}$

$\text{Using formula, }a^2-b^2=(a+b)(a-b)$

$\Rightarrow \dfrac{(\sec\theta+\tan\theta)(\sec\theta+1-\tan\theta)}{\sec\theta+1-\tan\theta}$

$\Rightarrow \sec\theta+\tan\theta$

$\sec\theta=\dfrac{1}{\cos\theta}\text{ and }\tan\theta=\dfrac{\sin\theta}{\cos\theta}$

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

$\Rightarrow \dfrac{1+\sin\theta}{\cos\theta}=RHS=\dfrac{1+\sin\theta}{\cos\theta}$

LHS=RHS

Hence Proved

Answer 2

Answer:

Hope it helps you.....