Question

IS LHS=RHS ?

$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\sec A+\tan A=\frac{1+\sin A}{\cos A}=\sqrt{\frac{1+\sin A}{1-\sin A}}$

Say true or false.

• A. True
• B. False
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is A True

Simplifying, we get
$=\frac{tanA+secA-1}{tanA-secA+1}$
$=\frac{sinA+1-cosA}{sinA-1+cosA}$
$=\frac{sinA+(1-cosA)}{sin-(1-cosA)}$
$=\frac{(sinA+1-cosA{)}^2}{si{n}^{2}A-(1-cosA{)}^2}$
$=\frac{si{n}^{2}A+co{s}^2+1-2sinAcosA-2cosA+2sinA}{(1-co{s}^{2}A)-(1-cosA{)}^2}$
$=\frac{2-2sinAcosA-2cosA+2sinA}{(1-cosA)(1+cosA-1+cosA)}$
$=\frac{2(1-cosA)+2sinA(1-cosA)}{(1-cosA)\left(2cosA\right)}$
$=secA+tanA$
$=\frac{1+sinA}{cosA}$
$=\sqrt{\frac{(1+sinA{)}^2}{1-si{n}^{2}A}}$
$=\sqrt{\frac{1+sinA}{1-sinA}}$