Question

Prove that
=
Sec 0 + Tan 0-1
Tan 6 - Sec 0+1
Cos 0
1 - Sin 0

Answer 1

Step-by-step explanation:

the required answer is in the attatchment.hope it helps u..

Answer 2

proved

Step-by-step explanation:

LHS = $\frac{\sec\theta+\tan\theta-1}{\tan\theta-\sec \theta+1 }$

$\frac{\sec\theta+\tan\theta-1}{\tan\theta-\sec \theta+1 }\\\\\frac{(\sec\theta-\tan \theta)- (\sec^2\theta-\tan^2\theta)}{\tan\theta-\sec\theta+1}\\\\\frac{(\sec\theta+\tan \theta)- (\sec\theta + \tan\theta)(\sec\theta-\tan \theta)}{\tan\theta-\sec\theta+1}\\\\\frac{(\sec\theta + \tan\theta)(1- (\sec\theta - \tan\theta))}{\tan\theta-\sec\theta+1}$

$\frac{(\sec\theta + \tan\theta)(1-\sec\theta + \tan\theta)}{1-\sec\theta + \tan\theta}\\\\\sec\theta + \tan\theta\\\\\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}\\\\\frac{1+\sin\theta}{\cos\theta}\\\\\frac{(1+\sin\theta)(1-\sin\theta)}{\cos\theta(1-\sin\theta)}\\\\\frac{1-\sin^2\theta}{\cos\theta(1-\sin\theta}\\\\\frac{\cos^2\theta}{\cos\theta(1-\sin\theta}\\\\\frac{\cos\theta}{1-\sin\theta}$

hence proved

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