Question
If $\sin^2\theta\cos^2\theta(1+\tan^2\theta)(1+\cos^2\theta)=\lambda,$ then find is the value of $\lambda$.
✓
Answer
Given,
$\sin^2\theta\cos^2\theta(1+\tan^2\theta)(1+\cos^2\theta)=\lambda,$
$\Rightarrow\ \sin^2\theta\cos^2\theta\sec^2\theta\text{ cosec}^2\theta=\lambda$
$\Rightarrow\ (\sin^2\theta\text{ cosec}^2\theta)\times(\cos^2\theta\sec^2\theta)=\lambda$
$\Rightarrow\ \Big(\sin^2\theta\times\frac{1}{\sin^2\theta}\Big)\Big(\cos^2\theta\times\frac{1}{\cos^2\theta}\Big)=\lambda$
$\Rightarrow\ \lambda=1\times1=1$
Hence, the value of $\lambda$ is 1.
$\sin^2\theta\cos^2\theta(1+\tan^2\theta)(1+\cos^2\theta)=\lambda,$
$\Rightarrow\ \sin^2\theta\cos^2\theta\sec^2\theta\text{ cosec}^2\theta=\lambda$
$\Rightarrow\ (\sin^2\theta\text{ cosec}^2\theta)\times(\cos^2\theta\sec^2\theta)=\lambda$
$\Rightarrow\ \Big(\sin^2\theta\times\frac{1}{\sin^2\theta}\Big)\Big(\cos^2\theta\times\frac{1}{\cos^2\theta}\Big)=\lambda$
$\Rightarrow\ \lambda=1\times1=1$
Hence, the value of $\lambda$ is 1.
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