Question

Heres my question

Trigonometry

Class 10

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Answer 1

SOLUTION

Refer to the attachment.

hope it helps ☺️⬆️

Answer 2

$\dfrac{tan \: \theta}{1 \: - \: cot \: \theta} \: + \: \dfrac{cot \: \theta}{1 \: - \: tan \: \theta} \: = \: 1 \: + \: sec \: \theta \: cosec \theta$

__________ [ GIVEN ]

• We have to prove L.H.S. = R.H.S.

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=> $\dfrac{tan \: \theta}{1 \: - \: cot \: \theta} \: + \: \dfrac{cot \: \theta}{1 \: - \: tan \: \theta}$

=> $\dfrac{ \frac{sin \theta}{cos \theta} }{1 \: - \: \frac{cos \theta}{sin \theta} } \: + \: \dfrac{\frac{cos \theta}{sin \theta}}{1 \: - \: \frac{sin \theta}{cos \theta} }$

=> $\dfrac{ \frac{sin \theta}{cos \theta} }{\frac{sin \: \theta \: - \: cos \theta}{sin \theta} } \: + \: \dfrac{\frac{cos \theta}{sin \theta}}{ \frac{cos \: \theta \: - \: sin \theta}{cos \theta} }$

=> $\dfrac{sin \theta}{cos \theta} \: \times \: \dfrac{sin \theta}{sin \theta \: - \: cos \theta} \: + \: \dfrac{cos \theta}{sin \theta} \: \times \: \dfrac{cos \theta }{cos \theta \: - \: sin \theta}$

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

=> $\dfrac{1}{sin \theta \: - \: cos \theta} \bigg( \dfrac{ {sin}^{3} \theta \: - \: {cos}^{3} \theta}{sin \theta \: cos \theta} \bigg)$

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

=> $\dfrac{1\: + \: sin \theta \: + \: cos \theta }{sin \theta \: cos \theta}$

=> $\dfrac{1 }{sin \theta \: cos \theta}$ + $\dfrac{sin \theta \: cos \theta }{sin \theta \: cos \theta}$

=> $1 \: + \: sec \: \theta \: cosec \theta$

L.H.S. = R.H.S.

_________ [ HENCE PROVED ]

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✡ Formula used :

tanØ = sinØ/cosØ

cotØ = cosØ/sinØ

sinØ = 1/secØ

cosØ = 1/cosecØ

a³ - b³ = (a - b) (a² + b² + ab)

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