Question

if 3x=cosecA &3/x=cotA then find 3[x (square )-1/x(square )]

Answer 1

3x = cosecA. 3/x = cotA
x = cosecA/3 x = 3/cotA
x² = cosec²A/9 1/x = cotA/3
1/x² = cot²A/9
3[x²-1/x²]= 3[cosec²A/9 - cot²A/9]
=3/9 (cosec²A-cot²A)
= 1/3

Answer 2

Answer:

The value of 3( x² -$\frac{1}{x^2}$ ) =  $\frac{1}{3}$

Step-by-step explanation:

Given,

cosec A = 3x

cot A = $\frac{3}{x}$

To find

The value of $3(x^2 - \frac{1}{x^2} )$

Solution:

Recall the concept

Cosec²θ - cot²θ = 1

We have,

Since cosec A = 3x,

we have x = $\frac{1}{3}$ cosecA

squaring on both sides,

x² =  $\frac{1}{9}$cosec²A -------------(1)

Also, cot A = $\frac{3}{x}$

$\frac{1}{x}$ = $\frac{1}{3}cotA$

Squaring on both sides

$\frac{1}{x^2}$ = $\frac{1}{9} \ cot^2A$ --------------------(2)

Subtracting equation (2) from (1) we get

(1) - (2) →,

x² -$\frac{1}{x^2}$  =   $\frac{1}{9}$ cosec²A - $\frac{1}{9}$ cot ²A

= $\frac{1}{9}$ (cosec²A - cot²A )

= $\frac{1}{9}$ (∵ By using the identity cosec²A - cot²A  = 1)

3( x² -$\frac{1}{x^2}$  ) = 3× $\frac{1}{9}$ =  $\frac{1}{3}$

The value of 3( x² -$\frac{1}{x^2}$ ) =  $\frac{1}{3}$

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