Question 45

What is the simplified value of $$(\frac{1}{secA+tanA})^2$$ ?

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)

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