Question

STATEMENT - 1 : $\frac{1+sinA}{1-sinA}=secA-tanA$
STATEMENT - 2 : $\frac{sinA-2si{n}^{2}A}{2co{s}^{3}A-cosA}=tanA$

• A. Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
• B. Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1
• C. Statement - 1 is True, Statement - 2 is False
• D. Statement - 1 is False, Statement - 2 is True

Answer 1

The correct option is D Statement - 1 is False, Statement - 2 is True

$\sqrt{\frac{1+sinA}{1-sinA}}=\sqrt{\frac{1+sinA}{1-sinA}\times \frac{1+sinA}{1+sinA}}$
$=\sqrt{\frac{(1+sinA{)}^2}{1-si{n}^{2}A}}=\frac{1+sinA}{\sqrt{co{s}^{2}A}}=\frac{1+sinA}{cosA}=secA+tanA$

$\frac{sinA-2si{n}^{2}A}{2co{s}^{3}A-cosA}=\frac{sinA(1-2si{n}^{2}A)}{cosA(2co{s}^{2}A-1)}=tanA\frac{cos2A}{cos2A}=tanA$