Question

Given A is an acute angle and cosec A =root over 2 find the value of 2sin^2 + 3cot^2/tan^2 A - cos^2 A

Answer 1

A is acute and cosec A = √2
⇒ A = 45°

$\frac{2\ sin^2A+3\ cot^2A}{tan^2A-cos^2A}\\ \\ = \frac{2\ sin^245^0+3\ cot^245^0}{tan^245^0-cos^245^0}\\ \\ = \frac{2 \times (1/ \sqrt{2})^2 +3 \times (1)^2}{(1)^2-(1/ \sqrt{2})^2 }\\ \\ = \frac{2 \times (1/2)+3}{1-(1/2)}\\\\= \frac{1+3}{1/2}\\\\= \frac{4}{1/2}\\\\=\boxed{8}$

Answer 2

The value of $\frac{2sin^{2}A+3cot^{2}A}{tan^{2}A-cos^{2}A}$ is $8$

Step-by-step explanation:

Given,

A is an acute angle and $cosecA=\sqrt{2}$

Find the value of $\frac{2sin^{2}A+3cot^{2}A}{tan^{2}A-cos^{2}A}=?$

∴ $cosecA=cosec45$

⇒$A=45$

∴ $\frac{2sin^{2}A+3cot^{2}A}{tan^{2}A-cos^{2}A}$

Plug the value of $A$,

$=\frac{2sin^{2}45+3cot^{2}45}{tan^{2}45-cos^{2}45}$

$=\frac{2(\frac{1}{\sqrt{2} }) ^{2}+3(1)^{2}}{(1)^{2}-(\frac{1}{\sqrt{2} }) ^{2}}$

$=\frac{2\times \frac{1}{2}+3 }{1-\frac{1}{2} }$

$=\frac{1+3}{\frac{1}{2} }$

$=\frac{4}{\frac{1}{2} }$

$=8$

So, The value of $\frac{2sin^{2}A+3cot^{2}A}{tan^{2}A-cos^{2}A}$ is $8$