Question

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Answer 1

Answer:

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Answer 2

QuEsTiOn :-

Prove that

$\frac{ {cos}^{2} \theta }{ {sin}^{2}\theta } - cosec\theta + sin\theta = 0$

Trigonometry formula used :-

$\implies {sin}^{2} \theta + {cos}^{2} \theta = 1 \\ \implies{cos}^{2} \theta = 1 - {sin}^{2} \theta \\ \implies \: cosec\theta = \frac{1}{sin \theta}$

SoLuTiOn :-

Taking LHS =>

$\implies\dfrac{ {cos}^{2} \theta}{ sin\theta } - cosec\theta + sin\theta \\ \\ \implies \dfrac{1 - {sin}^{2} \theta}{sin\theta} - \dfrac{1}{sin\theta} + sin\theta \\ \\ Taking \: LCM \\ \\ \implies \dfrac{1 - {sin}^{2} \theta - 1 + {sin}^{2}\theta }{sin\theta} \\ \\ \implies \dfrac{0}{sin\theta} \\ \\ \implies0 = RHS$

Here ,

${\purple{\boxed{\large{\bold{LHS = RHS}}}}}$

Hence Proved ✓