Math, asked by SamyekShakya, 11 months ago

Please solve the following identities

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Answers

Answered by phalgunagopal
1

Answer:

simple man use the identity

 {a}^{2}  -  {b}^{2}  = ( a + b) \times (a  - b)

in the numerator.

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Answered by Anonymous
125

AnswEr :

To Prove :

 \sf \dfrac{1 -  { \sin}^{4} ( \theta)}{ {  \cos}^{4} ( \theta)}  = 1 + 2 { \tan }^{2}( \theta)

Proof :

\leadsto \rm \dfrac{1 -  { \sin}^{4} ( \theta)}{ {  \cos}^{4} ( \theta)}

\leadsto \rm \dfrac{(1)^{2}  -  ({ \sin}^{2}( \theta))^{2} }{ {  \cos}^{4} ( \theta)}

⠀⠀⠀⠀⋆ (a² - b²) = (a + b)(a - b)

\leadsto \rm \dfrac{(1  +   { \sin}^{2} ( \theta))(1 -  { \sin}^{2} ( \theta))}{ {  \cos}^{4} ( \theta)}

⠀⠀⠀⠀⋆ (1 - sin² θ = cos² θ)

\leadsto \rm \dfrac{(1  +   { \sin}^{2} ( \theta))  \cancel{{\cos}^{2} ( \theta)}}{  \cancel{{  \cos}^{4} ( \theta)}}

\leadsto \rm \dfrac{(1  +   { \sin}^{2} ( \theta))}{{  \cos}^{2} ( \theta)}

\leadsto \rm\dfrac{1}{{  \cos}^{2} ( \theta)} +  \dfrac{{ \sin}^{2} ( \theta)}{{  \cos}^{2} ( \theta)}

⠀⠀⠀⠀⋆ 1 / cos² θ = sec² θ

⠀⠀⠀⠀⋆ sin² θ / cos² θ = tan² θ

\leadsto \rm \sec^{2} ( \theta)  +  \tan^{2} ( \theta)

⠀⠀⠀⠀⋆ sec² θ = (1 + tan² θ)

\leadsto \rm 1 + \tan^{2} ( \theta)   +  \tan^{2} ( \theta)

 \leadsto  \large\rm 1 + 2\tan^{2} ( \theta)

 \therefore \boxed{ \sf \dfrac{1 -  { \sin}^{4} ( \theta)}{ {  \cos}^{4} ( \theta)}  = 1 + 2 { \tan }^{2}( \theta) }

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