Question

If $A = 110^\circ $, then prove that \[\dfrac{{1 + \sqrt {1 + {{\tan }^2}2A} }}{{\tan 2A}} = - \tan A\].

Answer 1

Hint: We can take the LHS of the equation and simplify the terms inside the roots using trigonometric identities. Then we can change the signs of the trigonometric functions at given angles by checking which quadrant the angle lies. Then we can again simplify using appropriate identities to obtain the RHS. Thus, we can prove by saying that LHS is equal to the RHS.

Complete step-by-step answer:
We are given the value of A as, $A = 110^\circ $
Therefore, 2 times A will become,
 $ \Rightarrow 2A = 2 \times 110^\circ $
 $ \Rightarrow 2A = 220^\circ $
So, this angle lies in the \[{3^{rd}}\] quadrant as $220^\circ > 180^\circ $.
We need to prove that \[\dfrac{{1 + \sqrt {1 + {{\tan }^2}2A} }}{{\tan 2A}} = - \tan A\]. So, we can take the LHS.
 \[ \Rightarrow LHS = \dfrac{{1 + \sqrt {1 + {{\tan }^2}2A} }}{{\tan 2A}}\]
We know that ${\sec ^2}x = 1 + {\tan ^2}x$. So, the LHS will become,
 \[ \Rightarrow LHS = \dfrac{{1 + \sqrt {{{\sec }^2}2A} }}{{\tan 2A}}\]
On cancelling the square and square root, we get,
 \[ \Rightarrow LHS = \dfrac{{1 + \sec 2A}}{{\tan 2A}}\]
We know that sec is negative in the \[{3^{rd}}\] quadrant and tan is positive.
 \[ \Rightarrow LHS = \dfrac{{1 - \sec 2A}}{{\tan 2A}}\]
Now we can write the trigonometric ratios in terms of sin and cos. We know that $\tan x = \dfrac{{sinx}}{{\cos x}}$ and $\sec x = \dfrac{1}{{\cos x}}$. So, the LHS will become,
 \[ \Rightarrow LHS = \dfrac{{1 - \dfrac{1}{{\cos 2A}}}}{{\dfrac{{\sin 2A}}{{\cos 2A}}}}\]
On simplification, we get,
 \[ \Rightarrow LHS = \dfrac{{\cos 2A - 1}}{{\sin 2A}}\]
We know that \[\cos 2A = 1 - 2{\sin ^2}A\] and \[\sin 2A = 2\sin A\cos A\]. On substituting these in the LHS, we get
 \[ \Rightarrow LHS = \dfrac{{1 - 2{{\sin }^2}A - 1}}{{2\sin A\cos A}}\]
On simplification, we get,
 \[ \Rightarrow LHS = \dfrac{{ - 2{{\sin }^2}A}}{{2\sin A\cos A}}\]
On cancelling the common terms, we get,
 \[ \Rightarrow LHS = \dfrac{{ - \sin A}}{{\cos A}}\]
Since, $\tan x = \dfrac{{sinx}}{{\cos x}}$ so we have,
 \[ \Rightarrow LHS = - \tan A\]
But we have $RHS = - \tan A$. So, we can write,
 $ \Rightarrow LHS = RHS$
Thus, the required equation is proved.
Note: The trigonometric identities used in this problem are,
 ${\sec ^2}x = 1 + {\tan ^2}x$
 $\tan x = \dfrac{{sinx}}{{\cos x}}$
 $\sec x = \dfrac{1}{{\cos x}}$
We must know the sign of the trigonometric ratios in its respective quadrant. In the \[{1^{st}}\] quadrant all the trigonometric ratios are positive. In the \[{2^{nd}}\] quadrant only sine and sec are positive. In the third quadrant, only tan and cot are positive and in the fourth quadrant, only cos and sec are positive. The following figure gives us an idea about the signs of different trigonometric functions. The angle measured in the counter clockwise direction is taken as positive and angle measured in the clockwise direction is taken as negative.