Question

Prove that ${{\sin }^{2}}\dfrac{\pi }{6}+{{\cos }^{2}}\dfrac{\pi }{3}-{{\tan }^{2}}\dfrac{\pi }{4}=-\dfrac{1}{2}$

Answer 1

Hint: Use the fact that $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2},\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$ and $\tan \left( \dfrac{\pi }{4} \right)=1$. Substitute these values in the expression and simplify and hence prove L.H.S. = R.H.S.
Complete step-by-step answer:
Trigonometric ratios:
There are six trigonometric ratios defined on an angle of a right-angled triangle which are sine, cosine, tangent, cotangent, secant and cosecant.
The sine of an angle is defined as the ratio of the opposite side to the hypotenuse.
The cosine of an angle is defined as the ratio of the adjacent side to the hypotenuse.
The tangent of an angle is defined as the ratio of the opposite side to the adjacent side.
The cotangent of an angle is defined as the ratio of the adjacent side to the opposite side.
The secant of an angle is defined as the ratio of the hypotenuse to the adjacent side.
The cosecant of an angle is defined as the ratio of the hypotenuse to the opposite side.
Observe that sine and cosecant are multiplicative inverses of each other, cosine and secant are multiplicative inverses of each other, and tangent and cotangent are multiplicative inverses of each other.
We have the following table for the trigonometric ratios of $0,\dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{3},\dfrac{\pi }{2}$

From the above table, we have
$\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2},\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$ and $\tan \left( \dfrac{\pi }{4} \right)=1$
Hence, we have
${{\sin }^{2}}\dfrac{\pi }{6}+{{\cos }^{2}}\dfrac{\pi }{3}-{{\tan }^{2}}\dfrac{\pi }{4}={{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{1}{2} \right)}^{2}}-1=-\dfrac{1}{2}$
Hence, we have
L.H.S = R.H.S
Q.E.D
Note: [1] Many students do not remember the above table in radians, but remember it in degrees.
A simple conversion for radians into degrees is by replacing $\pi $ to $180{}^\circ $.