Question

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

Answer 1

$$\begin{gathered} \cot \theta \tan \left(90°-\theta \right)-\sec \left(90°-\theta \right)\mathrm{cosec}\theta +{\sin }^{2}65°+{\sin }^{2}25°+\sqrt{3}\tan 5°\tan 45°\tan 85° \\ =\cot \theta \cot \theta -\sec \left(90°-\theta \right)\mathrm{cosec}\theta +{\sin }^{2}65°+{\sin }^{2}25°+\sqrt{3}\tan 5°\tan 45°\tan 85°\left(\because \tan \left(90°-\theta \right)=\cot \theta \right) \\ ={\cot }^2\theta -\mathrm{cosec}\theta \mathrm{cosec}\theta +{\sin }^{2}65°+{\sin }^{2}25°+\sqrt{3}\tan 5°\tan 45°\tan 85°\left(\because \sec \left(90°-\theta \right)=\mathrm{cosec}\theta \right) \\ ={\cot }^2\theta -{\mathrm{cosec}}^2\theta +{\left(\sin \left(90°-25°\right)\right)}^2+{\sin }^{2}25°+\sqrt{3}\tan 5°\tan 45°\tan 85° \\ ={\cot }^2\theta -{\mathrm{cosec}}^2\theta +{\cos }^{2}25°+{\sin }^{2}25°+\sqrt{3}\tan 5°\tan 45°\tan 85°\left(\because \sin \left(90°-\theta \right)=\cos \theta \right) \\ =-1+{\cos }^{2}25°+{\sin }^{2}25°+\sqrt{3}\tan 5°\tan 45°\tan 85°\left(\mathrm{using}\mathrm{the}\mathrm{identity}:{\mathrm{cosec}}^2\theta -{\cot }^2\theta =1\right) \\ =-1+1+\sqrt{3}\tan 5°\tan 45°\tan 85°\left(\mathrm{using}\mathrm{the}\mathrm{identity}:{\cos }^2\theta +{\sin }^2\theta =1\right) \\ =0+\sqrt{3}\tan 5°\tan 45°\tan \left(90°-5°\right) \\ =\sqrt{3}\tan 5°\tan 45°\cot 5°\left(\because \tan \left(90°-\theta \right)=\cot \theta \right) \\ =\sqrt{3}\tan 5°\tan 45°\frac{1}{\tan 5°}\left(\because \cot \theta =\frac{1}{\tan \theta }\right) \\ =\sqrt{3}\tan 45° \\ =\sqrt{3}\left(\because \tan 45°=1\right) \\ \mathrm{Hence},\cot \theta \tan \left(90°-\theta \right)-\sec \left(90°-\theta \right)\mathrm{cosec}\theta +{\sin }^{2}65°+{\sin }^{2}25°+\sqrt{3}\tan 5°\tan 45°\tan 85°=\sqrt{3}. \end{gathered}$$