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+1 vote
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in Mathematics by (30 points)
\( \cot \theta \cdot \tan \left(90^{\circ}-\theta\right)-\sec \left(90^{\circ}-\theta\right) \operatorname{cosec} \theta+ \) \( \left(\sin ^{2} 25^{\circ}+\sin ^{2} 65\right)+\sqrt{3}\left(\begin{array}{l}\tan 5^{\circ} \tan 15^{\circ} \tan 30^{\circ} \\ \tan 75^{\circ} \tan 85^{\circ}\end{array}\right) \)

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1 Answer

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 We have, 

\(\cot \theta \cdot \tan \left(90^{\circ}-\theta\right)-\sec \left(90^{\circ}-\theta\right) \operatorname{cosec} \theta+ \) \(\left(\sin ^{2} 25^{\circ}+\sin ^{2} 65\right)+\sqrt{3}\left(\begin{array}{l}\tan 5^{\circ} \tan 15^{\circ} \tan 30^{\circ} \\ \tan 75^{\circ} \tan 85^{\circ}\end{array}\right) \)

= cotθ.cotθ - cosecθ.cosecθ + sin225 + cos225 + \(\sqrt{3}\)(tan5° tan15° tan30° cot15° cot5°)

= cot2θ - cosec2θ + 1 + \(\sqrt{3}\)(\(\frac{1}{3}\))

= -1 + 1 + \(\sqrt{3}\)

\(\sqrt{3}\)

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