Question

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Answer 1

To prove:

$\star \: \frac{ { \cos}^{2} \theta \: + { \tan}^{2} \theta - 1 }{ { \sin}^{2} \theta} = { \tan}^{2} \theta$

Solution:

LHS

$\implies\: \frac{ { \cos}^{2} \theta \: + { \tan}^{2} \theta - 1 }{ { \sin}^{2} \theta} \\ \ \\ \implies \: \frac{ { \cos}^{2} \theta \: + { \tan}^{2} \theta - ( { \sin}^{2} \theta + { \cos}^{2 } \theta }{ { \sin}^{2} \theta} \\ \\ \implies \: \frac{ { \cos}^{2} \theta \: + { \tan}^{2} \theta - { \sin}^{2} \theta - { \cos}^{2 } \theta }{ { \sin}^{2} \theta} \\ \\ \implies \: \frac{ { \tan}^{2} \theta - { \sin}^{2} \theta}{ { \sin}^{2} \theta} \\ \\ \implies \: \frac{ \frac{ { \sin}^{2} \theta }{ { \cos}^{2} \theta } - { \sin}^{2} \theta }{ { \sin}^{2} \theta} \\ \\ \implies \: \frac{ { \sin}^{2} \theta - { \sin}^{2} \theta \: { \cos}^{2} \theta}{ { \sin}^{2} \theta \: { \cos}^{2} \theta } \\ \\ \implies \: \frac{ { \cancel{\sin}^{2} \theta }(1 - { \cos}^{2} \theta)}{ \cancel{ { \sin}^{2} \theta} { \cos}^{2} \theta } \\ \\ \: \\ \\ \implies \: \frac{ { \sin}^{2} \theta }{ { \cos}^{2} \theta} \\ \\ \implies \: { \tan}^{2} \theta$

LHS = RHS .

$\huge\boxed{ \mathfrak{ \red{ \: proved}}}$

Formula used:

$\star \: { \sin}^{2} \alpha + { \cos}^{2} \alpha = 1 \\ \\ \star \: \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

Answer 2

Step-by-step explanation:

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