Question

please solve this question​

Answer 1

Consider the LHS.

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

Divide both the numerator and the denominator by $\cos\theta.$

$\dfrac {\left (\dfrac{1+\cos\theta+\sin\theta}{\cos\theta}\right)}{\left (\dfrac {1+\cos\theta-\sin\theta}{\cos\theta}\right)}\\\\\\=\dfrac {1+\sec\theta+\tan\theta}{1+\sec\theta-\tan\theta}$

Then multiply both by $\sec\theta-\tan\theta.$ So,

$\dfrac {(1+\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(1+\sec\theta-\tan\theta)(\sec\theta-\tan\theta)}\\\\\\=\dfrac {1+\sec\theta-\tan\theta}{(1+\sec\theta-\tan\theta)(\sec\theta-\tan\theta)}\\\\\\=\dfrac {1}{\sec\theta-\tan\theta}\\\\\\=\dfrac {1}{\left (\dfrac {1-\sin\theta}{\cos\theta}\right)}\\\\\\=\dfrac {\cos\theta}{1-\sin\theta}$

Now multiply both by $1+\sin\theta.$ Thus,

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

Hence we've arrived at the RHS.

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Answer 2

Step-by-step explanation:

please see photo in step by step explanation