Solution
Expression : $$(\frac{1}{secA+tanA})^2$$
= $$(\frac{}{\frac{1}{cosA}+\frac{sinA}{cosA}})^2$$
= $$(\frac{cosA}{1+sinA})^2=\frac{cos^2A}{(1+sinA)^2}$$
= $$\frac{1-sin^2A}{(1+sinA)^2}=\frac{(1-sinA)(1+sinA)}{(1+sinA)^2}$$
= $$\frac{1-sinA}{1+sinA}$$
=> Ans - (C)
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
