Question

Expression $\frac{tan \: A}{1 - cot \: A} + \frac{cot \: A}{1 - tan \: A}$ can be written :

(a) sinA cosA + 1

(b) secA cosecA + 1

(c) tanA + cotA

(d) secA + cosecA




$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \small \: {@swarnimkumar22}$

Answer 1

hey mate your answer is here ✌✌

so option b is the right answer

Answer 2

HELLO DEAR,

GIVEN:- tanA/(1 - cotA) + cotA/(1 - tanA)

$\bold{\implies \frac{sinA/cosA}{1 - cosA/sinA} + \frac{cosA/sinA}{1 - sinA/cosA}}$

$\bold{\implies \frac{sin^2A}{cosA(sinA - cosA)} + \frac{cos^2A}{sinA(cosA - sinA)}}$

$\bold{\implies \frac{sin^2A}{cosA(sinA - cosA)} - \frac{cos^2A}{sinA(sinA - cosA)}}$

$\bold{\implies \frac{sin^3A - cos^3A}{sinA.cosA(sinA - cos)}}$

$\bold{\implies \frac{\cancel{(sinA - cosA)}(sin^2A + cos^2A + sinAcosA)}{sinA.cosA\cancel{(sinA - cosA)}}}$
[as, a³ - b³ = (a - b)(a² + b² + ab) & (sin²θ + cos²θ = 1]

$\bold{\implies \frac{(1 + sinA.cosA)}{sinA.cosA}}$

$\bold{\implies \frac{1}{sinA.cosA} + \frac{sinA.cosA}{sinA.cosA}}$

$\bold{\implies secA.cosecA + 1}$

hence, option (b) is correct

$\bold{--------------OR--------------}$

$\bold{\implies \frac{tanA}{(1 - 1/tanA)} + \frac{(1/tanA)}{(1 - tanA)}}$

$\bold{\implies \frac{tan^2A}{(tanA - 1)} + \frac{1}{tanA(1 - tanA)}}$

$\bold{\implies \frac{1}{tan(1 - tanA)} - \frac{tan^2A}{(1 - tanA)}}$

$\bold{\implies \frac{1 - tan^3A}{tanA(1 - tanA)}}$

$\bold{\implies \frac{\cancel{(1 - tanA)}\{(1 + tan^2A) + tanA\}}{tanA\cancel{(1 - tanA)}}}$

$\bold{\implies \frac{sec^2A + tanA}{tanA}}$

$\bold{\implies \frac{1}{cos^2A}\frac{cosA}{sinA} + \frac{tanA}{tanA}}$

$\bold{\implies secA.cosecA + 1}$

hence, option (b) is correct

I HOPE IT'S HELP YOU DEAR,
THANKS