Question

If $\tan \theta -\cot \theta =a,\sin \theta +\cos \theta =b$, then $(b^2-1{)}^2(a^2+4)$ is equal to
$(\theta \ne \frac{(2n+1)\pi }{2}and\theta n\pi n\in I)$

• A. $2$
• B. $-4$
• C. $-2$
• D. $4$

Answer 1

Answer: (D)

$4$

$\begin{array}{c}Giventheequation........\\ sin\theta +cos\theta =b\\ squaringbothside:\\ \Rightarrow {\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =b^2|\begin{array}{c}{\sin }^2\theta +{\cos }^2\theta =1\\ 2\sin \theta \cos \theta =\sin 2\theta \end{array}\\ \Rightarrow \sin 2\theta =b^2-1--------\left(i\right)\\ Now,\\ tan\theta -cot\theta =a\\ \Rightarrow \frac{\sin \theta }{\cos \theta }-\frac{\cos \theta }{\sin \theta }=a\\ \Rightarrow \frac{{\sin }^2\theta -{\cos }^2\theta }{\sin \theta \cos \theta }=a\\ Now,multiplyby2:\\ \Rightarrow \frac{-2({\cos }^2\theta -{\sin }^2\theta )}{2\sin \theta \cos \theta }=a\\ \Rightarrow \frac{-2\cos 2\theta }{\sin 2\theta }=a\\ Now,squaringbothside:\\ \Rightarrow {\left(\frac{-2\cos 2\theta }{\sin 2\theta }\right)}^2=a^2\\ \Rightarrow \left(\frac{4{\cos }^{2}2\theta }{{\sin }^{2}2\theta }\right)=a^2\\ Again,adding4bothside:\\ \Rightarrow 4+\frac{4{\cos }^{2}2\theta }{{\sin }^{2}2\theta }=a^2+4\\ \Rightarrow 4\left[\frac{{\sin }^{2}2\theta +{\cos }^{2}2\theta }{{\sin }^{2}2\theta }\right]=a^2+4\\ \Rightarrow \frac{4}{{\sin }^{2}2\theta }=a^2+4----------\left(ii\right)\\ Now,findthevalueof:\\ wehave,\\ \Rightarrow (b^2-1{)}^2(a^2+4)=\left({\sin }^{2}2\theta \right)\left(\frac{4}{{\sin }^{2}2\theta }\right)\\ \therefore (b^2-1{)}^2(a^2+4)=4\\ sothatthecorrectoptionisD.\end{array}$