Question

tan θ +sin θ ÷tan θ -sin θ =sec θ +1÷sec θ -1
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Answer 1

Hello !

Question:-

$\dfrac{tanA+SinA}{tanA-SinA}$ = $\dfrac{SecA+1}{secA-1}$

given:-

• L.H.S = $\dfrac{tanA+SinA}{tanA-SinA}$

• R.H.S = $\dfrac{SecA+1}{secA-1}$

Proof :-

Refer to the attachment.

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Answer 2

Step-by-step explanation:

Hello !

Question:-

\dfrac{tanA+SinA}{tanA-SinA}

tanA−SinA

tanA+SinA

= \dfrac{SecA+1}{secA-1}

secA−1

SecA+1

given:-

L.H.S = \dfrac{tanA+SinA}{tanA-SinA}

tanA−SinA

tanA+SinA

R.H.S = \dfrac{SecA+1}{secA-1}

secA−1

SecA+1

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