Given A is an acute angle and cosec A = √2, find the value of
2 \sin ^{2} A+3 \cot ^{2} A /
\left(\tan ^{2} A-\cos ^{2} A\right)
The values of all trigonometric functions dependent on the value of the ratio of sides in a right-angled triangle are known as trigonometric ratios. The trigonometric ratios of a right-angled triangle's sides with regard to any of its acute angles are known as that angle's trigonometric ratios.
Let triangle ABC be a right-angled at B and A is an acute angle.
Given that cosec A = √2
Which implies,
AC/BC = √2/1
Let AC = √2x
Then BC = x
In right angled triangle ABC
By using Pythagoras theorem,
We get
\begin{array}{l} A C^{2}=A B^{2}+B C^{2} \\ (\sqrt{2} x)^{2}=A B^{2}+x^{2} \\ A B^{2}=2 x^{2}-x^{2} \\ A B=x \end{array}
Sin A = perpendicular/ hypotenuse
= BC/AC
= 1/ √2
Cot A = base/ perpendicular
= x/x
= 1
Tan A = perpendicular/ base
= BC/AB
= x/x
= 1
Cos A = base/ hypotenuse
= AB/AC
= x/ √2x
= 1/√2
Substituting these values we get
2 \sin ^{2} A+3 \cot ^{2} A /\left(\tan ^{2} A-\cos ^{2} A\right)=8
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