Question

Sec A (1- Sih A) (SecA+ tan A) ² 1​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{sec(A)\{1-sin(A)\}\{sec(A)+tan(A)\}}$

$\sf{=\dfrac{1}{cos(A)}\cdot\{1-sin(A)\}\bigg\{\dfrac{1}{cos(A)}+\dfrac{sin(A)}{cos(A)}\bigg\}}$

$\sf{=\dfrac{1}{cos(A)}\cdot\{1-sin(A)\}\bigg\{\dfrac{1+sin(A)}{cos(A)}\bigg\}}$

$\sf{=\dfrac{1}{cos(A)}\cdot\bigg\{\dfrac{(1-sin(A))(1+sin(A))}{cos(A)}\bigg\}}$

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

$\sf{=\dfrac{cos^2(A)}{cos^2(A)}}$

$\sf{=1}$