Question

1√7-√6
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Answer 1

Question:

$\bf \implies \: \dfrac{ { \tan( \theta) }^{2} }{ \sec( \theta - 1)^{2} } = \dfrac{1 + \cos( \theta) }{1 - \cos( \theta) }$

LHS:

$\implies \sf \: \dfrac{ \dfrac{ {sin}^{2} \theta }{ {cos}^{2} \theta } }{ \big(\dfrac{1}{cos \theta} - 1 \big)^{2} } = \dfrac{ \dfrac{ {sin}^{2} \theta }{ {cos}^{2} \theta } }{ \dfrac{(1 - cos \theta)^{2} }{cos ^{2} } \theta}$

$\implies \sf \: \dfrac{ {sin}^{2} \theta }{1 + {cos}^{2} \theta \: - 2 \: cos \theta}$

RHS:

$\implies \sf \dfrac{1 + cos \theta}{1 - cos \theta} \times \dfrac{1 - cos \theta}{1 - cos \theta}$

$\implies \sf \: \dfrac{1 - cos^{2} \theta}{(1 - cos \theta)^{2} } = \dfrac{sin^{2} \theta}{1 + {cos}^{2} \theta - 2 cos \theta}$

LHS = RHS, hence proved!!

Extras:

Trigonometry is basically, the study of shapes.

$\dfrac{OPPOSITE}{HYPOTENUSE} = sin$

$\dfrac{ADJACENT}{HYPOTENUSE} = cos$

$\dfrac{OPPOSITE}{ADJACENT} = tan$

$\dfrac{HYPOTENUSE}{OPPOSITE} = cosec$

$\dfrac{HYPOTENUSE}{ADJACENT} = sec$

$\dfrac{ADJACENT}{OPPOSITE} = cot$