Question

Prove the identities:

√[1+sinA/1-sinA] = sec A + tan A


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Answer 1

LHS=[1+sinA/1-SinA]^1/2

multiply with 1+SinA on both numerator and denominator

LHS=[(1+sinA)(1+SInA)/(1-sinA)(1+sinA)]^1/2

LHS=[(1+SinA)^2/cos^2]^1/2

LHS=(1+sinA)/cosA

LHS=1/cosA +(1/sinA)

LHS=secA+TanA

Answer 2

To Prove :

$\tt \sqrt{ \frac{1 + \sin A}{1 - \sin A} } = \sec A + \tan A.$

Concepts Required :

• (a + b )( a+ b) = (a + b)².
• 1/ Cos theta = Sec theta.
• Sin theta / Cos theta = tan theta.
• Sin² theta + Cos² theta = 1.
• 1+ tan² theta = Sec² theta.

Proof :

Taking LHS,

$\tt \dashrightarrow\sqrt{ \frac{1 + \sin A}{1 - \sin A} }$

$\tt \dashrightarrow\sqrt{ \frac{1 + \sin A}{1 - \sin A} } \times \sqrt{ \frac{1 + \sin A}{1 + \sin A} }$

$\tt \dashrightarrow\sqrt{ \frac{{(1 + \sin A) }^{2} }{1 - {\sin }^{2} A} }$

$\tt \dashrightarrow\sqrt{ \frac{{(1 + \sin A) }^{2} }{{\cos }^{2} A} }$

Now, Remove Root by square.

$\tt \dashrightarrow \frac{1 + \sin A}{ \cos A }$

Or, We can also write as.

$\tt \dashrightarrow \frac{1 }{ \cos A } + \frac{ \sin A}{ \cos A }$

$\tt \dashrightarrow \sec A + \tan A.$

Hance Proved.