Question : If $a(\tan\theta +\cot \theta)=1$, $\sin\theta +\cos\theta =b$ with $0^{\circ}< \theta < 90^{\circ}$, then a relation between $a$ and $b$ is:
Option 1: $b^{2}=2\left ( a+1 \right )$
Option 2: $b^{2}=2\left ( a-1 \right )$
Option 3: $2a=b^{2}-1$
Option 4: $2a=b^{2}+1$
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Correct Answer: $2a=b^{2}-1$
Solution :
$a(\tan\theta +\cot \theta) = 1$
$⇒a(\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}) = 1$
$⇒a(\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}) = 1$
$⇒a={\sin\theta\cos\theta}$
$\sin\theta +\cos\theta = b$ ......(i)
Squaring both sides,
$(\sin\theta +\cos\theta)^2 = b^2$
$⇒(\sin^2\theta +\cos^2\theta+2\sin\theta\cos\theta) = b^2$
$⇒1+2\sin\theta\cos\theta = b^2$
From equation (i)
$⇒1+2a = b^2$
$⇒2a=b^{2}-1$
Hence, the correct answer is $2a=b^{2}-1$.
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