Question

Prove :
=1 + 2 tan2 0 - 2 tan 0 sec O
1+ sin e ​

Answer 1

$\large \bf \orange{ \underline{ \underline{Solution \: : }}} \\ \\ \\ \bf \odot \: \: \: \: \: \: \: \: \: \: \: \frac{1 - \sin( \theta) }{1 + \sin( \theta) } \\ \\ \\ \implies \: \tt \: \frac{1 - \sin( \theta) }{1 + \sin( \theta) } \times \frac{1 - \sin( \theta) }{1 - \sin( \theta) } \\ \\ \implies \tt \: \: \frac{1 + { \sin}^{2}( \theta) - 2 \sin( \theta) }{{ \cos}^{2} ( \theta) } \\ \\ \: \: \: \: \mathcal{ \big[ \: using \: \: \bf{ \underline {{\sin }^{2} ( theta)+{ \cos ( theta)}^{2} = 1 }}\big]} \\ \\ \\ \implies \tt \: { \sec}^{2} ( \theta) + {\tan }^{2} ( \theta) - 2 \tan( \theta) \sec( \theta) \\ \\ \\ \big[ \: \: using \: \: { \underline{\bf{ \sec }^{2} ( theta) - { \tan }^{2} ( theta) = 1 }}\: \big]\\ \\ \\ \implies \: \:\green { \frak{ 1 + 2 {\tan }^{2} ( \theta) - 2 \tan( \theta) \sec( \theta)}} \: \: \: \: \: \: \: \: \bf \: proved.$

Answer 2

$\textbf{\huge{\red{Question}} }$

Prove : $1 + 2 tan2 0 - 2 tan 0 sec O$

$\textbf{\huge{\blue{Solution}} }$

$\implies$ $\dfrac{1 - sin(θ)}{1 + sin(θ)}$ × $\dfrac{1 - sin(θ)}{1 - sin(θ)}$

$\implies$ $\dfrac{1 + sin²(θ) - 2sin(θ)}{cos²(θ)}$

$\implies$ [using sin²(theta) + cos(theta)² = 1]

$\implies$ $sec²(θ) + tan²(θ) - 2 tan (θ) sec(θ)$

$\implies$ $[ using\:sec²(theta) - tan²(theta) = 1 ]$

$\implies$ $1 + 2 tan²(θ) - 2 tan(θ) sec(θ)$

$\implies$$\large\underline\orange{Hence\: Proved \:!!}$