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.