Question

Find the value of this expression $\dfrac{2\sin {{39}^{\circ }}}{\cos {{51}^{\circ }}}-\dfrac{\sec {{46}^{\circ }}}{\text{cosec}{{44}^{\circ }}}$.
\[\begin{align}
  & A.1 \\
 & B.2 \\
 & C.3 \\
 & D.-1 \\
\end{align}\]

Answer 1

Hint: In this question, we need to find the value of $\dfrac{2\sin {{39}^{\circ }}}{\cos {{51}^{\circ }}}-\dfrac{\sec {{46}^{\circ }}}{\text{cosec}{{44}^{\circ }}}$. For this we will use properties of trigonometric which are given as:
1: \[\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta \] similarly we can say that \[\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta \].
2: \[\sec \left( {{90}^{\circ }}-\theta \right)=\text{cosec}\theta \] similarly we can say that \[\text{cosec}\left( {{90}^{\circ }}-\theta \right)=\text{sec}\theta \].
Using these formulas will cancel out all trigonometric functions and we will be able to find the required value.

Complete step-by-step answer:
Here we are given the expression as $\dfrac{2\sin {{39}^{\circ }}}{\cos {{51}^{\circ }}}-\dfrac{\sec {{46}^{\circ }}}{\text{cosec}{{44}^{\circ }}}$.
We need to find its value. For this, we should simplify this expression using a trigonometric formula.
As we can see, ${{39}^{\circ }}+{{51}^{\circ }}$ equals ${{90}^{\circ }}$ so it means we can write ${{39}^{\circ }}$ as ${{90}^{\circ }}+{{51}^{\circ }}$ also.
Hence our expression becomes $\dfrac{2\sin \left( {{90}^{\circ }}-{{51}^{\circ }} \right)}{\cos {{51}^{\circ }}}-\dfrac{\sec {{46}^{\circ }}}{\text{cosec}{{44}^{\circ }}}$.
From trigonometry, we know that ${{90}^{\circ }}-\theta $ lies in I quadrant and the formula for \[\sin \left( {{90}^{\circ }}-\theta \right)\] is \[\cos \theta \]. So our expression becomes $\dfrac{2\cos {{51}^{\circ }}}{\cos {{51}^{\circ }}}-\dfrac{\sec {{46}^{\circ }}}{\text{cosec}{{44}^{\circ }}}$.
Cancelling $\cos {{51}^{\circ }}$ in the first term we get: $2-\dfrac{\sec {{46}^{\circ }}}{\text{cosec}{{44}^{\circ }}}$.
Now ${{46}^{\circ }}+{{44}^{\circ }}$ is also ${{90}^{\circ }}$. So we can write ${{46}^{\circ }}$ as ${{90}^{\circ }}-{{44}^{\circ }}$ we get: $2-\dfrac{\sec \left( {{90}^{\circ }}-{{44}^{\circ }} \right)}{\text{cosec}{{44}^{\circ }}}$.
As we know from trigonometry that \[\sec \left( {{90}^{\circ }}-\theta \right)=\text{cosec}\theta \] with positive sign because ${{90}^{\circ }}-\theta $ lies in first quadrant where all trigonometric functions are positive so we get: $2-\dfrac{\text{cosec}{{44}^{\circ }}}{\text{cosec}{{44}^{\circ }}}$.
Cancelling $\text{cosec}{{44}^{\circ }}$ we get: $2-1$.
Hence the value of our expression becomes equal to 1.

So, the correct answer is “Option A”.

Note: Students should keep in mind about signs while changing the quadrant. In the first quadrant, all the trigonometric functions are positive. In the second quadrant, only the sine and cosecant functions are positive. In the third quadrant, only the cosine and secant functions are positive. In the fourth quadrant, only the tangent and cotangent functions are positive. Students can also convert denominator as equal to the numerator and then cancel out as $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta \text{ and cosec}\left( {{90}^{\circ }}-\theta \right)=\text{sec}\theta $.