Question

Prove that $3\sin \dfrac{\pi }{6}\sec \dfrac{\pi }{3}-4\sin \dfrac{5\pi }{6}\cot \dfrac{\pi }{4}=1$ .

Answer 1

Hint: Try to simplify the left-hand side of the equation that we need to prove by using the formula $\sin \left( \pi -x \right)=\sin x$ , followed by using the required values of the trigonometric ratios using the trigonometric table of standard angles.

Complete step-by-step answer:

Here we will solve the left-hand side of the equation given in the question.
$3\sin \dfrac{\pi }{6}\sec \dfrac{\pi }{3}-4\sin \dfrac{5\pi }{6}\cot \dfrac{\pi }{4}$
$=3\sin \dfrac{\pi }{6}\sec \left( \dfrac{\pi }{2}-\dfrac{\pi }{6} \right)-4\sin \left( \pi -\dfrac{\pi }{6} \right)\cot \dfrac{\pi }{4}$
Now we know that $\sin \left( \pi -x \right)=\sin x$ and $\sec \left( \dfrac{\pi }{2}-\dfrac{\pi }{6} \right)=\cos ec\dfrac{\pi }{6}$ .
$3\sin \dfrac{\pi }{6}\cos ec\dfrac{\pi }{6}-4\sin \dfrac{\pi }{6}\cot \dfrac{\pi }{4}$
We know that $\dfrac{1}{\sin x}=\cos ecx$ . Therefore, our expression becomes:
$3\sin \dfrac{\pi }{6}\times \dfrac{1}{\sin \dfrac{\pi }{6}}-4\sin \dfrac{\pi }{6}\cot \dfrac{\pi }{4}$
$=3-4\sin \dfrac{\pi }{6}\cot \dfrac{\pi }{4}$
Now we know that $\sin \dfrac{\pi }{6}=\dfrac{1}{2}\text{ and }\cot \dfrac{\pi }{4}=1$ .
$3-4\times \dfrac{1}{2}\times 1$
=1
As we have shown that the left-hand side of the equation given in the question is equal to the right-hand side of the equation. So, we can say we have proved that $3\sin \dfrac{\pi }{6}\sec \dfrac{\pi }{3}-4\sin \dfrac{5\pi }{6}\cot \dfrac{\pi }{4}=1$ .

Note: Be careful about the calculation and the signs while opening the brackets. The general mistake that a student can make is 1+x-(x-1)=1+x-x-1. Also, you need to remember the properties related to complementary angles and trigonometric ratios. The above question could also be solved by putting the values of the required trigonometric ratios using the trigonometric table in the first step itself, but this would require a good knowledge of signs of the values of trigonometric ratios in different quadrants.