Solve the following

Question:

Show that $\frac{\cos ^{2}\left(45^{\circ}+\theta\right)+\cos ^{2}\left(45^{\circ}-\theta\right)}{\tan \left(60^{\circ}+\theta\right) \tan \left(30^{\circ}-\theta\right)}=1$

 

Solution:

$\mathrm{LHS}=\frac{\cos ^{2}\left(45^{\circ}+\theta\right)+\cos ^{2}\left(45^{\circ}-\theta\right)}{\tan \left(60^{\circ}+\theta\right) \cdot \tan \left(30^{\circ}-\theta\right)}$

$=\frac{\cos ^{2}\left(45^{\circ}+\theta\right)+\left[\sin \left\{90^{\circ}-\left(45^{\circ}-\theta\right)\right\}\right]^{2}}{\tan \left(60^{\circ}+\theta\right) \cdot \cot \left\{90^{\circ}-\left(30^{\circ}-\theta\right)\right\}}$

$\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right.$ and $\left.\cot \left(90^{\circ}-\theta\right)=\tan \theta\right]$

$=\frac{\cos ^{2}\left(45^{\circ}+\theta\right)+\sin ^{2}\left(45^{\circ}+\theta\right)}{\tan \left(60^{\circ}+\theta\right) \cdot \cot \left(60^{\circ}+\theta\right)}$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$=\frac{1}{\tan \left(60^{\circ}+\theta\right) \cdot \frac{1}{\tan \left(60^{\circ}+\theta\right)}}=1=\mathrm{RHS}$ $[\because \cot \theta=1 / \tan \theta]$

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