Question

Without using trigonometric tables, evaluate the following:
$${\displaystyle \frac{\sec 37{}^{\circ }}{\text{cosec }53{}^{\circ }}}+2\cot 15{}^{\circ }\cot 25{}^{\circ }\cot 45{}^{\circ }\cot 75{}^{\circ }\cot 65{}^{\circ }-3({\sin }^{2}18{}^{\circ }+{\sin }^{2}72{}^{\circ })$$

Answer 1

Hint: Use trigonometric identities to solve this question. Use the formula for conversion of sec and cosec to cos and sin respectively.
 $$\begin{array}{rl}& \text{cosec}(\theta )={\displaystyle \frac{1}{\text{sin}(\theta )}}\\ & \sec (\theta )={\displaystyle \frac{1}{\text{cos}(\theta )}}\end{array}$$

Use the formula of complementary angles in trigonometry like
$$\sin (\theta )=\cos ({90}^{\circ }-\theta )$$
$$\tan ({90}^{\circ }-\theta )=\cot (\theta )$$

Use the formula
$${\sin }^2(\theta )+{\cos }^2(\theta )=1$$

Complete step-by-step answer:
This is a very basic question on trigonometric identities. In this question many trigonometric identities are to be used.
We know that,
$$\begin{array}{rl}& \text{cosec}(\theta )={\displaystyle \frac{1}{\text{sin}(\theta )}}...\text{(i)}\\ & \sec (\theta )={\displaystyle \frac{1}{\text{cos}(\theta )}}...\text{(ii)}\end{array}$$
Using the equation (i) and equation (ii) we get the first term converted as:
$${\displaystyle \frac{\sec 37{}^{\circ }}{\text{cosec }53{}^{\circ }}}={\displaystyle \frac{{\displaystyle \frac{1}{\cos 37{}^{\circ }}}}{{\displaystyle \frac{1}{\text{sin }53{}^{\circ }}}}}={\displaystyle \frac{\sin 53{}^{\circ }}{\cos 37{}^{\circ }}}...\text{(iii) }$$
We know that
 $$\sin (\theta )=\cos ({90}^{\circ }-\theta )$$

Now using the above equation in the numerator of right hand side of equation (iii) we get,
$${\displaystyle \frac{\sin 53{}^{\circ }}{\cos 37{}^{\circ }}}={\displaystyle \frac{\cos ({90}^{\circ }-53{}^{\circ })}{\cos 37{}^{\circ }}}={\displaystyle \frac{\cos 37{}^{\circ }}{\cos 37{}^{\circ }}}=1$$
So,
$${\displaystyle \frac{\sec 37{}^{\circ }}{\text{cosec }53{}^{\circ }}}=1...\text{(iv)}$$
We know that
$$\tan ({90}^{\circ }-\theta )=\cot (\theta )$$

Now using the above equation we get,
$$\cot 15{}^{\circ }=\tan ({90}^{\circ }-15{}^{\circ })=\tan 75{}^{\circ }$$
Similarly we can get
$$\cot 25{}^{\circ }=\tan ({90}^{\circ }-25{}^{\circ })=\tan 65{}^{\circ }$$

Also, we know that
$$\cot 45{}^{\circ }=1$$
Replacing $$\cot 15{}^{\circ }$$, $$\cot 25{}^{\circ }$$and $$\cot 45{}^{\circ }$$ with the expression as found above in the second term we get the second term as:
$$2\cot 15{}^{\circ }\cot 25{}^{\circ }\cot 45{}^{\circ }\cot 75{}^{\circ }\cot 65{}^{\circ }=2\tan 75{}^{\circ }\tan 65{}^{\circ }\cdot 1\cdot \cot 75{}^{\circ }\cot 65{}^{\circ }$$

Rearranging the terms in right hand side of the equation we get,
$$2\cot 15{}^{\circ }\cot 25{}^{\circ }\cot 45{}^{\circ }\cot 75{}^{\circ }\cot 65{}^{\circ }=2\tan 75{}^{\circ }\cot 75{}^{\circ }\tan 65{}^{\circ }\cot 65{}^{\circ }...\text{(v)}$$

We know that
$$\begin{array}{rl}& \tan (\theta )={\displaystyle \frac{1}{\text{cot}(\theta )}}\\ & \Rightarrow \tan (\theta )\cot (\theta )=1...\text{(vi)}\end{array}$$

Using equation (vi) we get,
$$\tan 75{}^{\circ }\cot 75{}^{\circ }=1$$

Similarly, $$\tan 65{}^{\circ }\cot 65{}^{\circ }=1$$
Using the above expression in the equation (v) we get,
$$\begin{array}{rl}& 2\cot 15{}^{\circ }\cot 25{}^{\circ }\cot 45{}^{\circ }\cot 75{}^{\circ }\cot 65{}^{\circ }=2(\tan 75{}^{\circ }\cot 75{}^{\circ })(\tan 65{}^{\circ }\cot 65{}^{\circ })\\ & =2(1)(1)\\ & =2\end{array}$$
So, we get that
$$2\cot 15{}^{\circ }\cot 25{}^{\circ }\cot 45{}^{\circ }\cot 75{}^{\circ }\cot 65{}^{\circ }=2...\text{(vii)}$$

Now coming to the last term of the given expression,
We know that
$$\sin (\theta )=\cos ({90}^{\circ }-\theta )$$
Using the above equation we get,
 $$\begin{array}{rl}& \sin 72{}^{\circ }=\cos ({90}^{\circ }-72{}^{\circ })\\ & =\cos 18{}^{\circ }\end{array}$$

Using the above conversion, the third term is converted to:
 $$\begin{array}{rl}& -3({\sin }^{2}18{}^{\circ }+{\sin }^{2}72{}^{\circ })=-3({\sin }^{2}18{}^{\circ }+{(\sin 72{}^{\circ })}^2)\\ & =-3({\sin }^{2}18{}^{\circ }+{(\cos 18{}^{\circ })}^2)\\ & =-3({\sin }^{2}18{}^{\circ }+{\cos }^{2}18{}^{\circ })...\text{(viii)}\end{array}$$
We know that
$${\sin }^2(\theta )+{\cos }^2(\theta )=1$$
Using the above equation we get,
 $${\sin }^{2}18{}^{\circ }+{\cos }^{2}18{}^{\circ }=1...\text{(ix)}$$

Using equation (ix) in the equation (viii) we get,
$$\begin{array}{rl}& -3({\sin }^{2}18{}^{\circ }+{\sin }^{2}72{}^{\circ })=-3({\sin }^{2}18{}^{\circ }+{\cos }^{2}18{}^{\circ })\\ & =-3(1)\\ & =-3\end{array}$$
So, we get
$$-3({\sin }^{2}18{}^{\circ }+{\sin }^{2}72{}^{\circ })=-3...\text{(x)}$$

Now using equation (iii), equation (vii) and equation (x), adding them all we get,
$$\begin{array}{rl}& {\displaystyle \frac{\sec 37{}^{\circ }}{\text{cosec }53{}^{\circ }}}+2\cot 15{}^{\circ }\cot 25{}^{\circ }\cot 45{}^{\circ }\cot 75{}^{\circ }\cot 65{}^{\circ }-3({\sin }^{2}18{}^{\circ }+{\sin }^{2}72{}^{\circ })=1+2-3\\ & \Rightarrow {\displaystyle \frac{\sec 37{}^{\circ }}{\text{cosec }53{}^{\circ }}}+2\cot 15{}^{\circ }\cot 25{}^{\circ }\cot 45{}^{\circ }\cot 75{}^{\circ }\cot 65{}^{\circ }-3({\sin }^{2}18{}^{\circ }+{\sin }^{2}72{}^{\circ })=0\end{array}$$
Hence we get the value of the expression in the question as 0.

Note: This question is totally based on trigonometric identities so you should be well familiar with every identity otherwise you cannot do this question. There is one more way to simplify the second term, you can convert the next two factors into tan and then proceed to make a group of tan and cot to make it 1. There is a chance of omitting $$\cot 45{}^{\circ }$$ in the second term which makes the solution procedure wrong though this will not affect the answer.