Question

Evaluate:

${\left(\frac{3\sin \left(43\right)°}{\cos \left(47\right)°}\right)}^2-\frac{\cos \left(37\right)°\csc \left(53\right)°}{\tan \left(5\right)°\tan \left(25\right)°\tan \left(65\right)°\tan \left(85\right)°}$

Answer 1

Solution:

Given:

${\left(\frac{3\sin \left(43\right)°}{\cos \left(47\right)°}\right)}^2-\frac{\cos \left(37\right)°\csc \left(53\right)°}{\tan \left(5\right)°\tan \left(25\right)°\tan \left(65\right)°\tan \left(85\right)°}$

$={\left(\frac{3\sin \left(43\right)°}{\cos (90-43)°}\right)}^2-\frac{\cos \left(37\right)°\csc \left(90-37\right)°}{\tan \left(5\right)°\tan \left(25\right)°\tan \left(90-25\right)°\tan \left(90-5\right)°}$

$={\left(\frac{3\sin \left(43\right)°}{\sin \left(43\right)°}\right)}^2-\frac{\cos \left(37\right)°\sec \left(37\right)°}{\tan \left(5\right)°\tan \left(25\right)°\cot \left(25\right)°\cot \left(5\right)°}\mathbf{}\left[\because cos({90}^{°}-\theta )=sin\left(\theta \right),tan({90}^{°}-\theta )=cot\left(\theta \right)andcsc({90}^{°}-\theta )=sec\left(\theta \right)\right]$

$$\begin{gathered} ={\left(3\right)}^2-\frac{1}{1}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{t}\mathbf{a}\mathbf{n}\left(\theta \right)\mathbf{c}\mathbf{o}\mathbf{t}\left(\theta \right)\mathbf{=}\mathbf{1}\mathbf{}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{}\mathbf{s}\mathbf{e}\mathbf{c}\left(\theta \right)\mathbf{c}\mathbf{o}\mathbf{s}\left(\theta \right)\mathbf{=}\mathbf{1}\mathbf{]} \\ =9-1 \\ =8 \end{gathered}$$

Final Answer: Value of ${\left(\frac{3\sin \left(43\right)°}{\cos \left(47\right)°}\right)}^2-\frac{\cos \left(37\right)°\csc \left(53\right)°}{\tan \left(5\right)°\tan \left(25\right)°\tan \left(65\right)°\tan \left(85\right)°}\mathbf{}\mathbf{=}\mathbf{}\mathbf{8}$