Question

prove that tan(90-theta)+cot(90-theta)/cosec theta =sec A​

Answer 1

$\textbf{To prove:}$

$\mathsf{\dfrac{tan(90-\theta)+cot(90-\theta)}{cosec\theta}=sec\theta}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{\dfrac{tan(90-\theta)+cot(90-\theta)}{cosec\theta}}$

$\mathsf{=\dfrac{cot\theta+tan\theta}{cosec\theta}}$

$\mathsf{=\dfrac{\dfrac{cos\theta}{sin\theta}+\dfrac{sin\theta}{cos\theta}}{\dfrac{1}{sin\theta}}}$

$\mathsf{=\dfrac{\dfrac{cos^2\theta+sin^2\theta}{sin\theta\,cos\theta}}{\dfrac{1}{sin\theta}}}$

$\mathsf{=\dfrac{\dfrac{1}{sin\theta\,cos\theta}}{\dfrac{1}{sin\theta}}}$

$\mathsf{=\dfrac{1}{sin\theta\,cos\theta}{\times}\dfrac{sin\theta}{1}}$

$\mathsf{=\dfrac{1}{cos\theta}}$

$\mathsf{=sec\theta}$

$\implies\boxed{\mathsf{\dfrac{tan(90-\theta)+cot(90-\theta)}{cosec\theta}=sec\theta}}$

$\textbf{Find more:}$

Tan(π/4+x)-tan(π/4-x)=2tan 2x

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Answer 2

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