Question

Prove that (cosec theta-sintheta)(sec theta-cos theta)=(1)÷(tan theta +cot theta)

Answer 1

$\underline{\textbf{To prove:}}$

$\mathsf{(cosec\theta-sin\theta)(sec\theta-cos\theta)=\dfrac{1}{tan\theta+cot\theta}}$

$\underline{\textbf{Solution:}}$

$\textbf{L.H.S}$

$\mathsf{=(cosec\theta-sin\theta)(sec\theta-cos\theta)}$

$\mathsf{=\left(\dfrac{1}{sin\theta}-sin\theta\right)\left(\dfrac{1}{cos\theta}-cos\theta\right)}$

$\mathsf{=\left(\dfrac{1-sin^2\theta}{sin\theta}\right)\left(\dfrac{1-cos^2\theta}{cos\theta}\right)}$

$\mathsf{=\left(\dfrac{cos^2\theta}{sin\theta}\right)\left(\dfrac{sin^2\theta}{cos\theta}\right)}$

$\mathsf{=cos\theta\,sin\theta}$---------(1)

$\textbf{R.H.S}$

$\mathsf{\dfrac{1}{tan\theta+cot\theta}}$

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

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

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

$\mathsf{=cos\theta\;sin\theta}$ ------(2)

$\textsf{From (1) and (2)}$

$\textf{L.H.S=R.H.S}$

$\implies\boxed{\mathsf{(cosec\theta-sin\theta)(sec\theta-cos\theta)=\dfrac{1}{tan\theta+cot\theta}}}$

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