Question : If $\operatorname{cosec} A+\cot A=a \sqrt{b}$, then find the value of $\frac{\left(a^2 b-1\right)}{\left(a^2 b+1\right)}$.
Option 1: $\cos A$
Option 2: $\tan A$
Option 3: $\frac{1}{\sin A}$
Option 4: $\frac{1}{\cot A}$
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Correct Answer: $\cos A$
Solution :
Consider, $\operatorname{cosec} A+\cot A=a \sqrt{b}$
Squaring both sides, we get,
⇒ $(\operatorname{cosec} A+\cot A)^2=a^2b$
⇒ $\operatorname{cosec^2 }A+\cot^2A+2\operatorname{cosec} A\cot A = a^2b$
⇒ $a^2b=\frac{1}{\sin^2 A}+\frac{\cos^2A}{\sin^2 A}+2\frac{1}{\sin A}\frac{\cos A}{\sin A}$
⇒ $a^2b=\frac{1+\cos^2A+2\cos A}{\sin^2 A}$
⇒ $a^2b=\frac{(1+\cos A)^2}{\sin^2A}$
⇒ $a^2b=\frac{(1+\cos A)^2}{1-\cos^2A}$ [As $\cos^2A+\sin^2A=1$]
⇒ $a^2b=\frac{(1+\cos A)^2}{(1-\cos A)(1+\cos A)}$ [Using $a^2-b^2=(a-b)(a+b)$]
⇒ $\frac{a^2b}{1}=\frac{1+\cos A}{1-\cos A}$
Applying componendo and dividendo,
⇒ $\frac{a^2b-1}{a^2b+1}=\frac{(1+\cos A)-(1-\cos A)}{(1+\cos A)+(1-\cos A)}$
⇒ $\frac{a^2b-1}{a^2b+1}=\frac{1+\cos A-1+\cos A)}{2}$
⇒ $\frac{a^2b-1}{a^2b+1}=\frac{2\cos A}{2}$
⇒ $\frac{a^2b-1}{a^2b+1}=\cos A$
Hence, the correct answer is $\cos A$.
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