Question

prove that 1+cosA/1-cosA = tanA/(secA-1)​

Answer 1

Answer:

$\boxed{\purple{\Large{\mathsf{\frac{1 + cosA}{1 - cosA} = \frac{tanA}{(secA-1)}}}}}$

Step-by-step explanation:

To prove:

$\Large{\frac{1 + cosA}{1 - cosA} = \frac{tanA}{(secA-1)}}$

Proof:

L.H.S = $\Large{\frac{1 + cosA}{1 - cosA}}$

=> Dividing by cosA on both numerator and denominator, we obtain,

=> $\Large{\dfrac{\frac{1 + cosA}{cosA}}{\frac{1 - cosA}{cosA}}}$

=> $\Large{\dfrac{\frac{1}{cosA} + \frac{cosA}{cosA}}{\frac{1}{cosA} - \frac{cosA}{cosA}}}$

=> $\Large{\frac{secA + 1}{secA - 1}}$

$\Large{(\because{\frac{1}{cosA} = secA)}}$

=> By squaring

=> $\Large{\frac{sec^2A + 1^2}{(secA - 1)^2}}$

=> $\Large{\frac{tan^2A}{(secA - 1)^2}}$

[°.° 1 + tan²A = sec²A,

.°. tan²A = sec²A - 1]

Taking square root,

=> $\Large{\frac{tanA}{(secA - 1)}}$ = R.H.S

$\boxed{\pink{\Large{\mathsf{\frac{1 + cosA}{1 - cosA} = \frac{tanA}{(secA-1)}}}}}$