Question

Evaluate sec A cosec (90-A) - tan A cot (90-A) + sin^2 55 +sin 35 / tan 10 tan 20 tan 60 tan 70 tan 80​

Answer 1

$$\begin{gathered} \sec \mathrm{A}.\mathrm{cosec}\left(90°-\mathrm{A}\right)-\tan \mathrm{A}.\cot \left(90°-\mathrm{A}\right)+\frac{{\sin }^{2}55°+{\sin }^{2}35°}{\tan 10°.\tan 20°.\tan 60°.\tan 70°.\tan 80°} \\ =\sec \mathrm{A}.\sec \mathrm{A}-\tan \mathrm{A}.\tan \mathrm{A}+\frac{{\sin }^{2}55°+{\sin }^2\left(90°-55°\right)}{\tan 10°.\tan 20°.\sqrt{3}.\tan \left(90°-20°\right).\tan \left(90°-10°\right)} \\ ={\sec }^2\mathrm{A}-{\tan }^2\mathrm{A}+\frac{{\sin }^{2}55°+{\cos }^{2}55°}{\tan 10°.\tan 20°.\sqrt{3}.\cot 20°.\cot 10°} \\ =1+\frac{1}{\tan 10°.\tan 20°.\sqrt{3}.{\displaystyle \frac{1}{\tan 20°}}.{\displaystyle \frac{1}{\tan 10°}}} \\ =1+\frac{1}{\sqrt{3}} \\ =\frac{\sqrt{3}+1}{\sqrt{3}} \\ =\frac{\sqrt{3}+1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ =\frac{3+\sqrt{3}}{3} \end{gathered}$$