Question

tan+sin÷tan_sin= sec+1÷sec_1​

Answer 1

Step-by-step explanation:

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Answer 2

$\large\bf\underline{Question:}$

Prove that:

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

$\large\bf\underline{Solution:}$

$LHS = \frac{tan \theta+sin \theta }{tan \theta - sin \theta} \\ \implies\frac{ \frac{sin \theta}{cos \theta} + sin \theta}{ \frac{sin \theta}{cos \theta} - sin \theta } \\ \implies \frac{sin \theta( \frac{1}{cos \theta } + 1) }{sin \theta( \frac{1}{cos \theta } - 1)} \\ \implies \frac{\frac{1}{cos \theta } + 1}{\frac{1}{cos \theta } - 1} \\ \implies \frac{sec \theta + 1}{sec \theta - 1} = RHS \\ Hence \: Proved$

Identities used:

$1. \: tan \theta = \frac{sin \theta}{cos \theta} \\ 2. \: \frac{1}{cos \theta = sec \theta}$

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