Question

prove that 1+2 cos^2A / 1+3 cot^2 A = sin^2A

Answer 1

Step-by-step explanation:

(1+2cos²A)/1+3cot²A

=(1+cos^2A)sin^2/sin^2A+3cos^2 A

=(1+cos^2A)sin^2A/1+cos^2A

=sin^2A

Answer 2

Concept: Anytime trigonometric functions are used in an expression or equation, trigonometric Identities are helpful. For any value of a variable appearing on both sides of an equation, a trigonometric identity is true. These trigonometric functions (such as sine, cosine, and tangent) of one or more angles are involved geometrically in these identities.

Trigonometric Identities are equality statements that hold true for all values of the variables in the equation and that use trigonometry functions.

There are several distinctive trigonometric identities that relate a triangle's side length and angle. Only the right-angle triangle is consistent with the trigonometric identities.

All trigonometric identities are built upon the foundation of the six trigonometric ratios. Some of their names include sine, cosine, tangent, cosecant, secant, and cotangent. The adjacent side, opposite side, and hypotenuse side of the right triangle are used to determine each of these trigonometric ratios.The six trigonometric ratios are the source of all fundamental trigonometric identities.

Given: The equation $\dfrac{1+2\cos^2 \mbox{A}}{1+3\cot^2 \mbox{A}}= \sin^2 \mbox{A}$ is given

Find: We need to prove that the equation is true.

Solution:

$\mbox{L.H.S.}= \dfrac{1+2\cos^2 \mbox{A}}{1+3\cot^2 \mbox{A}}$

$=\dfrac{1+2\cos^2 \mbox{A}}{1+3\dfrac{\cos^2 \mbox{A}}{\sin^2 \mbox{A}}}$

$=\dfrac{1+2\cos^2 \mbox{A}}{\dfrac{\sin^2 \mbox{A}+3\cos^2 \mbox{A}}{\sin^2 \mbox{A}}}$

$=\dfrac{1+2\cos^2 \mbox{A}}{\frac{1+2\cos^2 \mbox{A}}{\sin^2 \mbox{A}}}~~~~~~ [\because \sin^2\mbox{x}+\cos^2\mbox{x}=1]$

$=\sin^2 \mbox{A}\left(\dfrac{1+2\cos^2 \mbox{A}}{1+2\cos^2 \mbox{A}}\right)$

= sin²A = R.H.S.

Hence, we prove L.H.S. = R.H.S

Therefore, $\dfrac{1+2\cos^2 \mbox{A}}{1+3\cot^2 \mbox{A}}= \sin^2\mbox{A}$ is proved.

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