Question

(sectheta-tantheta)square=1-sintheta/1+sintheta

Answer 1

Note

let ¢=x

To prove

$\sf (secx-tanx)^2=\dfrac{1-sinx}{1+sinx}$

Explanation:

Consider LHS

$= > {(secx - tanx)}^{2}$

Expanding using the formula

$\underline {\sf (a-b)^2=a^2+b^2-2ab}$

$= > {sec}^{2} x + {tan}^{2} x - 2secxtanx$

we know that

$\sf sec^2x-tan^2x=1$

$\underline{\sf sec^x=1+tan^2x}$

applying this formula

$= > 1 + {tan}^{2} x + {tan}^{2} x - 2secxtnx$

$= > 1 + 2 {tan}^{2} x - 2secxtanx$

$\underline{ \sf \: tanx = \dfrac{sinx}{cosx} }$

$\underline{ \sf \: secx = \dfrac{1}{cosx} }$

Using these

$= > 1 + 2 \dfrac{ {sin}^{2}x }{ {cos}^{2} x} - 2 \times \dfrac{1}{cosx} \times \dfrac{sinx}{cosx}$

$= > 1 + 2\dfrac{ {sin}^{2}x }{ {cos}^{2}x } - 2 \dfrac{sinx}{ {cos}^{2}x }$

Taking LCM

$= > \dfrac{ {cos}^{2}x + {sin}^{2} x + {sin}^{2} x - 2sinx }{ {cos}^{2} x}$

$\underline{ \sf \: {sin}^{2} x + {cos}^{2} x = 1}$

$= > \dfrac{ {sin}^{2} x - 2sinx + 1}{1 - {sin}^{2} x}$

Numerator is of the form (a-b)^2

$= > \dfrac{ {(1 - sinx)}^{2} }{ {1}^{2} - {sin}^{2} x}$

using

$\underline{ \sf \: {a}^{2} - {b}^{2} = (a + b)(a - b)}$

$= > \dfrac{(1 - sinx)(1 - sinx)}{(1 - sinx)(1 + sinx)}$

$= > \dfrac{1 - sinx}{1 + sinx}$

Hence proved...