Question

CosecA=4x+1/16x then cosecA+cotA

Answer 1

Answer:

$cosecA+cotA=8x$

Step-by-step explanation:

Concept used:

$\bold{cosec^2A-cot^2A=1}$

$\bold{(a+b)^2=a^2+b^2+2ab}$

$\bold{(a-b)^2=a^2+b^2-2ab}$

Given:

$cosecA=4x+\frac{1}{16x}$......(1)

squaring on both sides

$cosec^2A=(4x+\frac{1}{16x})^2$

$\implies\:cosec^2A=16x^2+\frac{1}{256x^2}+2*4x*\frac{1}{16x}$

$\implies\:cosec^2A=16x^2+\frac{1}{256x^2}+\frac{1}{2}$

$\implies\:cosec^2A-1=16x^2+\frac{1}{256x^2}+\frac{1}{2}-1$

$\implies\:cosec^2A-1=16x^2+\frac{1}{256x^2}-\frac{1}{2}$

$\implies\:cosec^2A-1=(4x-\frac{1}{16x})^2$

$\implies\:cot^2A=(4x-\frac{1}{16x})^2$

Taking square root on both sides

$\implies\:cotA=4x-\frac{1}{16x}$........(2)

Adding (1) and (2)

$cosecA+cotA=4x+\frac{1}{16x}+4x-\frac{1}{16x}$

$\implies\:cosecA+cotA=8x$