Question

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Answer 1

Correct Question :

$\tt \frac{ { \tan }^{2} \theta}{ {( \sec \theta - 1) }^{2} } = \frac{1 + \cos \theta }{1 + \cos \theta}$

Formula Required :

• 1+ tan² theta = Sec²theta.
• (a-b)(a+b) = a²- b².
• 1/Cos theta = Sec theta.

Solution :

Taking LHS,

$\dashrightarrow\tt\frac{ { \tan }^{2} \theta}{ {( \sec \theta - 1) }^{2} }$

$\dashrightarrow\tt \frac{ { \sec}^{2} \theta - 1}{ {( \sec \theta - 1) }^{2} }$

$\dashrightarrow\tt \frac{ (\sec \theta + 1)( \sec \theta - 1) }{( \sec \theta - 1) ( \sec \theta - 1)}$

$\dashrightarrow\tt \frac{ \sec \theta + 1}{ \sec \theta - 1}$

Now, Convert In Cos theta.

$\dashrightarrow\tt \frac{1 + \cos \theta }{1 - \cos \theta }$

# RHS = LHS

Hence Proved.

Answer 2

Question :-

To prove :-

$\frac{ {tan}^{2} \theta }{ ({sec \theta - 1)}^{2} } = \frac{1 + cos \theta}{1 - cos \theta}$

Trigonometry Identities Used :-

$1 + {tan}^{2} \theta = {sec}^{2} \theta \\ {tan}^{2} \theta = {sec}^{2} \theta - 1 \\(a - b)(a + b) = {a}^{2} - {b}^{2} \\ sec \theta = \frac{1}{cos \theta}$

Proof :-

Taking LHS

$\frac{ {tan}^{2} \theta}{( {sec \theta - 1)}^{2} } \\ \\ = \frac{ {sec}^{2} \theta - 1 }{ ({sec \theta - 1)}^{2} } \\ \\ = \frac{ {sec}^{2} \theta - {1}^{2} }{( {sec \theta - 1)}^{2} } \\ \\ = \frac{(sec \theta + 1)(sec \theta - 1)}{(sec \theta - 1)(sec \theta - 1)} \\ \\ = \frac{sec \theta + 1}{sec \theta - 1} \\ \\ = \frac{ \frac{1}{cos \theta} + 1 }{ \frac{1}{cos \theta} - 1 } \\ \\ = \frac{ \frac{1 + cos \theta}{cos \theta} }{ \frac{1 - cos \theta}{cos \theta} } \\ \\ = \frac{1 + cos \theta}{1 - cos \theta} = RHS \\ \\ {\purple{\boxed{\large{\bold{LHS = RHS}}}}}$

Hence proved