Question

sec A+tanA-1/tanA-sec +1 =cosA/1-sinA​

Answer 1

So we have to prove,

$\dfrac {\sec A+\tan A-1}{\tan A-\sec A+1}=\dfrac {\cos A}{1-\sin A}$

So, taking LHS,

$\dfrac {\sec A+\tan A-1}{\tan A-\sec A+1}$

By using the identity,

$\sec A+\tan A-1=(\sec A+\tan A)(\tan A-\sec A+1),$

the LHS becomes,

$\dfrac {(\sec A+\tan A)(\tan A-\sec A+1)}{\tan A-\sec A+1}\\\\\\=\sec A+\tan A\\\\\\=\dfrac {1}{\cos A}+\dfrac {\sin A}{\cos A}\\\\\\=\dfrac {1+\sin A}{\cos A}$

Multiplying both the numerator and the denominator by $1-\sin A,$

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

Thus we arrived at the RHS.

Hence the Proof!

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