Question

Prove that (tan A + sin A)/(tan A - sin A) = (sec A + 1)/(sec A - 1).

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Answer 1

$\huge\underline{\rm{☆To \ Prove}}$

$\rm{\dfrac{\ tanA + \ sinA}{\ tanA - \ sinA} = \dfrac{\ secA + 1}{\ secA - 1}}$

$\rm{LHS = \dfrac{\ tanA + \ sinA}{\ tanA - \ sinA}}$

$\rm{ = \dfrac{\dfrac{\ sinA}{\ cosA} + \ sinA}{\dfrac{\ sinA}{\ cosA} - \ sinA}}$

$\rm{ = \dfrac{ \dfrac{ \ sinA + \ sinA \ cosA }{ \ cosA }}{\dfrac{ \ sinA - \ sinA \ cosA}{ \ cosA}}}$

$\rm{ = \dfrac{\ sinA + \ sinA\ cosA}{\ sinA - \ sinA \ cosA}}$

$\rm{ = \dfrac{\ sinA(1 + \ cosA}{\ sinA(1 - \ cosA}}$

$\rm{ = \dfrac{\ sinA + \ dfrac{1}{\ secA}}{\ sinA - \dfrac{1}{\ secA}}}$

$\rm{ = \dfrac{\dfrac{\ secA + 1}{\ secA}}{\dfrac{\ secA - 1}{\ secA}}}$

$\rm{ = \dfrac{\ secA + 1}{\ secA - 1}= RHS}$

$\red{\boxed{\rm{Hence \ Proved✔}}}$

#NAWABZAADI

Answer 2

HeY TheRe!

$ \Large \mathcal \red{LHS :}$

$\sf{ \frac{(tan A + sin A)}{(tan A - sin A) }= \frac{ (sec A + 1)}{(sec A - 1)}}$

${\mapsto\sf{ \frac{ (tan A + sin A)}{(tan A – sin A)}}}$

${\mapsto \sf{\frac{ [ (\frac{sin A}{cos A}) + sin A]}{ [( \frac{sin A}{cos A}) – sin A]}}}$

${ \mapsto \sf{ \frac{[sin A({ \frac{1}{cos A}} + 1)]}{[sin A({ \frac{1}{cos A} }– 1)]}}}$

${ : \implies \bf \red{ \frac{ (sec A + 1)}{(sec A – 1)}}}$

$ \Large \mathcal \blue{: RHS}$

• Hence Proved...!!!!

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