Question

Prove that :::
-> sin theta/(sec theta+tan theta-1) + cos theta/(cosec theta+cot theta-1) = 1
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Answer 1

ANSWER:

To Prove:

$\:\:\:\bullet\:\:\:\dfrac{\sin\theta}{\sec\theta+\tan\theta-1}+\dfrac{\cos\theta}{\csc\theta+\cot\theta-1}=1$

Proof:

$\text{We need to prove that,}\\\\:\longrightarrow\dfrac{\sin\theta}{\sec\theta+\tan\theta-1}+\dfrac{\cos\theta}{\csc\theta+\cot\theta-1}=1\\\\\text{Taking LHS,}\\\\:\implies\dfrac{\sin\theta}{\sec\theta+\tan\theta-1}+\dfrac{\cos\theta}{\csc\theta+\cot\theta-1}\\\\\text{We know that,}\\\\:\hookrightarrow\sec\phi=\dfrac{1}{\cos\phi}\\\\:\hookrightarrow\tan\phi=\dfrac{\sin\phi}{\cos\phi}\\\\:\hookrightarrow\csc\phi=\dfrac{1}{\sin\phi}\\\\:\hookrightarrow\cot\phi=\dfrac{\cos\phi}{\sin\phi}\\\\\text{So,}$

$:\implies\dfrac{\sin\theta}{\sec\theta+\tan\theta-1}+\dfrac{\cos\theta}{\csc\theta+\cot\theta-1}\\\\:\implies\dfrac{\sin\theta}{\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}-1}+\dfrac{\cos\theta}{\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}-1}\\\\:\implies\dfrac{\sin\theta}{\frac{1+\sin\theta-\cos\theta}{\cos\theta}}+\dfrac{\cos\theta}{\frac{1+\cos\theta-\sin\theta}{\sin\theta}}$

$:\implies\dfrac{\sin\theta\times\cos\theta}{1+\sin\theta-\cos\theta}+\dfrac{\cos\theta\times\sin\theta}{1+\cos\theta-\sin\theta}\\\\\text{Taking (\sin\theta\cos\theta) common,}\\\\:\implies\sin\theta\cos\theta\left(\dfrac{1}{1+\sin\theta-\cos\theta}+\dfrac{1}{1+\cos\theta-\sin\theta}\right)\\\\\text{Taking LCM,}\\\\:\implies\sin\theta\cos\theta\left(\dfrac{1+\cos\theta-\sin\theta+1+\sin\theta-\cos\theta}{(1+\sin\theta-\cos\theta)(1+\cos\theta-\sin\theta)}\right)$

$:\implies\sin\theta\cos\theta\left(\dfrac{(1+1)+(\cos\theta-\cos\theta)+(\sin\theta-\sin\theta)}{(1+\cos\theta-\sin\theta+\sin\theta+\sin\theta\cos\theta-\sin^2\theta-\cos\theta-\cos^2\theta+\cos\theta\sin\theta)}\right)\\\\:\implies\sin\theta\cos\theta\left(\dfrac{2+0+0}{1+(\cos\theta-\cos\theta)+(\sin\theta-\sin\theta)+(\sin\theta\cos\theta+\cos\theta\sin\theta)-\sin^2\theta-\cos^2\theta}\right)$

$:\implies\sin\theta\cos\theta\left(\dfrac{2}{1+0+0+2\sin\theta\cos\theta-(\sin^2\theta+\cos^2\theta)}\right)\\\\\text{We know that,}\\\\:\hookrightarrow\sin^2\phi+\cos^2\phi=1\\\\\text{So,}$

$:\implies\sin\theta\cos\theta\left(\dfrac{2}{1+2\sin\theta\cos\theta-(\sin^2\theta+\cos^2\theta)}\right)\\\\:\implies\sin\theta\cos\theta\left(\dfrac{2}{1+2\sin\theta\cos\theta-1}\right)\\\\:\implies\sin\theta\cos\theta\left(\dfrac{2}{1-1+2\sin\theta\cos\theta}\right)\\\\:\implies\dfrac{2\sin\theta\cos\theta}{2\sin\theta\cos\theta}\\\\\bf{:\implies 1=RHS}\\\\\text{\bf{HENCE PROVED!!!}}$

Formulae Used:

$:\hookrightarrow1)\:\sec\phi=\dfrac{1}{\cos\phi}\\\\:\hookrightarrow2)\:\tan\phi=\dfrac{\sin\phi}{\cos\phi}\\\\:\hookrightarrow3)\:\csc\phi=\dfrac{1}{\sin\phi}\\\\:\hookrightarrow4)\:\cot\phi=\dfrac{\cos\phi}{\sin\phi}\\\\:\hookrightarrow5)\:\sin^2\phi+\cos^2\phi=1$