Question

(cose theatre + cot theatre) square=1+cos theatre/1-cos theatre​

Answer 1

CORRECT QUESTION :-

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

SOLUTION :-

$\\ \sf \: LHS = (cosec\theta - {cot\theta)}^{2} \\ \\ \boxed{ \bf \: cosec\theta = \dfrac{1}{sin\theta} } \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \:cot\theta = \dfrac{cos\theta}{sin\theta} }$

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

$\\ \boxed{ \bf \: {cos}^{2} + {sin}^{2} = 1 } \\ \boxed{ \bf \: {sin}^{2} = 1 - {cos}^{2} }$

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

$\\ \boxed{ \bf \: (a - {b)}^{2} = (a - b)(a - b)} \\ \\ \boxed{ \bf \: {a}^{2} - {b}^{2} = (a + b)(a - b)}$

$\\ \implies \sf \: \dfrac{ \cancel{(1 - cos\theta)}(1 - cos\theta)}{ \cancel{(1 - cos\theta)}(1 + cos\theta)} \\ \\ \implies \sf \: \dfrac{1 - cos\theta}{1 + cos\theta} = RHS \: \: \: \: (proved)$

MORE IDENTITIES :-

★ 1 + tan²ø = sec²ø

★ 1 + cot²ø = cosec²ø

★ sin2ø = 2.sinø.cosø

★ cos2ø = cos²ø - sin²ø

ㅤㅤㅤ ㅤ= 2cos²ø - 1

ㅤㅤ ㅤㅤ= 1 - 2sin²ø

Answer 2

Correct Question :

To Prove :

$\: \:\:\:\:\;\red\bigstar \: (Cosec\theta + cot\theta )^2 = \dfrac{1+cos\theta}{1-cos\theta}$

Solution :

$(Cosec\theta + cot\theta )^2 = \dfrac{1+cos\theta}{1-cos\theta}$

Solving R.H.S.

$\large \dfrac{1+cos\theta}{1-cos\theta}$

• Rationalising

$\large \dfrac{1+cos\theta}{1-cos\theta} \times \dfrac{1+cos\theta}{1+cos\theta}$

We know ,

$\red\bigstar \color{blue}\boxed{ (a+b)(a-b)=a^2-b^2 }$

Using this Identity,

$\large \dfrac{(1+cos\theta)^2}{1^2-cos^2 \theta}$

We know,

$\red\bigstar\boxed{ sin^2 \theta + cos^2 \theta = 1}$

So,

$\red\bigstar\boxed{ sin^2 = 1 - cos^2 \theta }$

Using this Identity , we got :

$\implies \large \dfrac{(1+cos\theta)^2}{sin ^2 \theta} \\ \\ \large \implies \displaystyle \left( \dfrac{1+cos\theta}{sin \theta} \right) ^2$

• Splitting

$\large \implies \displaystyle \left( \dfrac{1}{sin \theta}+ \dfrac{cos\theta}{sin \theta} \right) ^2$

• As we know,

• 1/sinθ = cosecθ
• sinθ/cosθ = cotθ

So,

$\large \implies \displaystyle \left( cosec\theta + cot\theta \right) ^2$

L.H.S = R.H.S

$\huge \underline{\underline{Hence \: \: proved }}$

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