Question

sin^2alpha / 1 + cot^2alpha + tan^2alpha / (1 + tan^2alpha)^2 + cos^2alpha = ?​

Answer 1

SOLUTION

TO DETERMINE

$\displaystyle \sf{ \frac{ { \sin}^{2} \alpha }{1 + { \cot}^{2} \alpha } + \frac{ { \tan}^{2} \alpha }{{(1 + { \tan}^{2} \alpha)}^{2} } + { \cos}^{2} \alpha \: }$

FORMULA TO BE IMPLEMENTED

We are aware of the Trigonometric identity that

$1. \: \: \: { \sin}^{2} \alpha + { \cos}^{2} \alpha = 1$

$2. \: \: \: \sf{ 1 + { \tan}^{2} \alpha = { \sec}^{2} \alpha }$

$3. \: \: \: \sf{ 1 + { \cot}^{2} \alpha = { \csc}^{2} \alpha}$

EVALUATION

$\displaystyle \sf{ \frac{ { \sin}^{2} \alpha }{1 + { \cot}^{2} \alpha } + \frac{ { \tan}^{2} \alpha }{{(1 + { \tan}^{2} \alpha)}^{2} } + { \cos}^{2} \alpha \: }$

$= \displaystyle \sf{ \frac{ { \sin}^{2} \alpha }{ { \csc}^{2} \alpha } + \frac{ { \tan}^{2} \alpha }{{ ( { \sec}^{2} \alpha)}^{2} } + { \cos}^{2} \alpha \: }$

$= \displaystyle \sf{ \frac{ { \sin}^{2} \alpha }{ \frac{1}{ { \sin}^{2} \alpha } } + \frac{ { \sin}^{2} \alpha }{ { \cos}^{2} \alpha } \times {( { \cos}^{2} \alpha)}^{2} + { \cos}^{2} \alpha \: }$

$= \displaystyle \sf{ { \sin}^{4} \alpha + ({ \sin}^{2} \alpha \times { \cos}^{2} \alpha ) + { \cos}^{2} \alpha \: }$

$= \displaystyle \sf{ { \sin}^{2} \alpha( { \sin}^{2} \alpha + { \cos}^{2} \alpha ) + { \cos}^{2} \alpha \: }$

$= \sf{ ({ \sin}^{2} \alpha \times 1) + { \cos}^{2} \alpha }$

$= \displaystyle \sf{ { \sin}^{2} \alpha+ { \cos}^{2} \alpha \: }$

$= 1$

Note :

In the above Solution csc means cosec

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Answer 2

Step-by-step explanation:

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