# Prove that

Question:

$\cot \theta \tan \left(90^{\circ}-\theta\right)-\sec \left(90^{\circ}-\theta\right) \operatorname{cosec} \theta+\sin ^{2} 65^{\circ}+\sin ^{2} 25^{\circ}+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ}$

Solution:

$\cot \theta \tan \left(90^{\circ}-\theta\right)-\sec \left(90^{\circ}-\theta\right) \operatorname{cosec} \theta+\sin ^{2} 65^{\circ}+\sin ^{2} 25^{\circ}+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ}$

$=\cot \theta \cot \theta-\sec \left(90^{\circ}-\theta\right) \operatorname{cosec} \theta+\sin ^{2} 65^{\circ}+\sin ^{2} 25^{\circ}+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ} \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right)$

$=\cot ^{2} \theta-\operatorname{cosec} \theta \operatorname{cosec} \theta+\sin ^{2} 65^{\circ}+\sin ^{2} 25^{\circ}+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ} \quad\left(\because \sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta\right)$

$=\cot ^{2} \theta-\operatorname{cosec}^{2} \theta+\left(\sin \left(90^{\circ}-25^{\circ}\right)\right)^{2}+\sin ^{2} 25^{\circ}+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ}$

$=\cot ^{2} \theta-\operatorname{cosec}^{2} \theta+\cos ^{2} 25^{\circ}+\sin ^{2} 25^{\circ}+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right)$

$=-1+\cos ^{2} 25^{\circ}+\sin ^{2} 25^{\circ}+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ} \quad$ (using the identity : $\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1$ )

$=-1+1+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ} \quad$ (using the identity : $\cos ^{2} \theta+\sin ^{2} \theta=1$ )

$=0+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan \left(90^{\circ}-5^{\circ}\right)$

$=\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \cot 5^{\circ} \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right)$

$=\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \frac{1}{\tan 5^{\circ}} \quad\left(\because \cot \theta=\frac{1}{\tan \theta}\right)$

$=\sqrt{3} \tan 45^{\circ}$

$=\sqrt{3} \quad\left(\because \tan 45^{\circ}=1\right)$

Hence, $\cot \theta \tan \left(90^{\circ}-\theta\right)-\sec \left(90^{\circ}-\theta\right) \operatorname{cosec} \theta+\sin ^{2} 65^{\circ}+\sin ^{2} 25^{\circ}+\sqrt{3} \tan 5^{\circ} \tan 45^{\circ} \tan 85^{\circ}=\sqrt{3}$.

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