Question

How do you simplify $\dfrac{{\sin x}}{{\csc x}} + \dfrac{{\cos x}}{{\sec x}}$ ?

Answer 1

Hint : We can solve the given problem if we know the reciprocals relationship of a trigonometric function. We know that the reciprocal of cosecant is sine and reciprocal of secant is cosine. Applying this to a given problem we obtained a simple expression. We have an identity which shows the relationship between sine and cosine function that is ${\sin ^2}x + {\cos ^2}x$

Complete step-by-step answer :
Given,
$\dfrac{{\sin x}}{{\csc x}} + \dfrac{{\cos x}}{{\sec x}}$ .
We can write this as
$= \dfrac{1}{{\csc x}} \times \sin x + \dfrac{1}{{\sec x}} \times \cos x$
We know that reciprocal of cosecant is sine and reciprocal of secant is cosine. That is
$\dfrac{1}{{\csc x}} = \sin x$ and $\dfrac{1}{{\sec x}} = \cos x$ .
Applying this to the given problem we have,
$= \left( {\sin x \times \sin x} \right) + \left( {\cos x \times \cos x} \right)$
$= {\sin ^2}x + {\cos ^2}x$
But we know the identity of a trigonometry. That is ${\sin ^2}x + {\cos ^2}x = 1$ .
Substituting this we have,
$= 1$
Thus we have,
$\dfrac{{\sin x}}{{\csc x}} + \dfrac{{\cos x}}{{\sec x}} = 1$ .
So, the correct answer is “1”.

Note : We have an identity which shows the relationship between tangent and secant. That is ${\tan ^2}x + 1 = {\sec ^2}x$
We have an identity which shows the relationship between cotangent and cosecant. That is
${\cot ^2}x + 1 = {\csc ^2}x$
Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions; sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively.