Question

How do you evaluate sec(40).sin(50)?

Answer 1

Here, in this problem, we need to evaluate the expression of trigonometric angles. We will use complementary angles, reciprocal identities and some basic computations. Complementary angles are those angles whose sum is $\dfrac{\pi }{2}$ i.e. ${{90}^{\circ }}$. Some formulae of complementary angles are:
$\begin{aligned} & \Rightarrow \sin \left( \dfrac{\pi }{2}-x \right)=\cos x \\\ & \Rightarrow \cos \left( \dfrac{\pi }{2}-x \right)=\sin x \\\ & \Rightarrow \tan \left( \dfrac{\pi }{2}-x \right)=\cot x \\\ & \Rightarrow \cot \left( \dfrac{\pi }{2}-x \right)=\tan x \\\ \end{aligned}$

Complete step by step answer:
Now, let’s begin to solve the question.
We all know about the basic trigonometric functions which are sin, cos and tan, and derived functions are those which are derived from the basic trigonometric functions like cot, sec and cosec.
All these functions are used to form the identities which are used to solve all the questions related to trigonometry. Let’s have a look at the reciprocal identities which we will need in this question.
$\Rightarrow$sinx = $\dfrac{1}{\cos ecx}$ or cosecx = $\dfrac{1}{\sin x}$
$\Rightarrow$cosx = $\dfrac{1}{\sec x}$ or secx = $\dfrac{1}{\cos x}$
$\Rightarrow$tanx = $\dfrac{\sin x}{\cos x}$ = $\dfrac{1}{\cot x}$
$\Rightarrow$cotx = $\dfrac{1}{\tan x}$ = $\dfrac{\cos x}{\sin x}$
There is a use of complementary angles to solve the question. Complementary angles are those angles whose sum is $\dfrac{\pi }{2}$ i.e. ${{90}^{\circ }}$. Let’s see them too.
$\begin{aligned} & \Rightarrow \sin \left( \dfrac{\pi }{2}-x \right)=\cos x \\\ & \Rightarrow \cos \left( \dfrac{\pi }{2}-x \right)=\sin x \\\ & \Rightarrow \tan \left( \dfrac{\pi }{2}-x \right)=\cot x \\\ & \Rightarrow \cot \left( \dfrac{\pi }{2}-x \right)=\tan x \\\ \end{aligned}$
Now write the expression given in the question.
$\Rightarrow$sec(40).sin(50)
As we know how secx is formed. We will apply reciprocal identity:
$\Rightarrow$cosx = $\dfrac{1}{\sec x}$ or secx = $\dfrac{1}{\cos x}$
Apply it:
$\Rightarrow \dfrac{\sin \left( 50 \right)}{\cos \left( 40 \right)}$
As we know that $\cos \left( \dfrac{\pi }{2}-x \right)=\sin x$. By applying this we will get:
$\Rightarrow \dfrac{\sin \left( 50 \right)}{\cos \left( \dfrac{\pi }{2}-40 \right)}$
We know 90-40 = 50:
$\Rightarrow \dfrac{\sin \left( 50 \right)}{\sin \left( 50 \right)}$
On cancelling the like terms we will get 1.
$\therefore$sec(40).sin(50) = 1

Note:
When we apply complementary angles, the sign remains the same here because it took place in the first quadrant where both x-axis and y-axis values are positive. The angle formed in the first quadrant is $\dfrac{\pi }{2}$. Do remember all the identities in trigonometry because these are the only sources which help in solving problems like this.