Question

can anyone solve this question
but shouldn't find LHS and RHS separately.
(cosec A-sin A)(sec A-cos A)=1÷tanA+cot A.

Answer 1

Step-by-step explanation:

$Given: (cosecA - sinA)(secA - cosA)$

$=(\frac{1}{sinA}-sinA)(\frac{1}{cosA}-cosA)$

$=(\frac{1 - sin^2A}{sinA})(\frac{1-cos^2A}{cosA})$

$=\frac{cos^2A}{sinA} * \frac{sin^2A}{cosA}$

$=sinA * cosA$

∴ sin²A + cos²A = 1

$=\frac{sinA*cosA}{sin^2A+cos^2A}$

$=\frac{1}{\frac{sin^2A+cos^2A}{sinA * cosA} }$

$=\frac{1}{\frac{sinA}{cosA} +\frac{cosA}{sinA}}$

$=\boxed{\frac{1}{tanA+cotA}}$

Hope it helps!

Answer 2

RHS: 1÷sinA÷cosA+cosA÷sinA

1÷sin^2A+cos^2A÷sinAcosA

1÷1÷sinAcosA

sinAcosA

LHS: (1÷sinA-sinA)(1÷cosA-cosA)

(1-sin^2A÷sinA)(1-cos^2A÷cosA)

(cos^A÷sinA)sin^2A÷cosA)

sinAcosA =RHS

I HOPE THIS WILL HELP YOU JUST WRITE IT AS I WROTE AND PLEASE DROP A THANKS AND VOTE ALSO

THANK YOU ······