Question

If the value of $ \sin x+\cos x=\dfrac{1}{5} $ , then find the value of $ \tan x $ ?

Answer 1

Hint: We start solving the problem by squaring the given equation on both sides. We then expand the square and then make use of the results $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ and $ 2\sin x\cos x=\sin 2x $ to proceed through the problem. We then make use of the result $ \sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x} $ and perform necessary calculations to get the quadratic equation in $ \tan x $ . We then factorize the quadratic equation and find the roots to get the required value(s) of $ \tan x $ .

Complete step by step answer:
According to the problem, we are given that $ \sin x+\cos x=\dfrac{1}{5} $ and we need to find the value of $ \tan x $ .
So, we have given $ \sin x+\cos x=\dfrac{1}{5} $ ---(1).
Let us square on both sides of the equation (1).
 $ \Rightarrow {{\left( \sin x+\cos x \right)}^{2}}={{\left( \dfrac{1}{5} \right)}^{2}} $ .
 $ \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x=\dfrac{1}{25} $ ---(2).
We know that $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ and $ 2\sin x\cos x=\sin 2x $ . Let us use this results in equation (2).
 $ \Rightarrow 1+\sin 2x=\dfrac{1}{25} $ .
 $ \Rightarrow \sin 2x=\dfrac{1}{25}-1 $ .
 $ \Rightarrow \sin 2x=\dfrac{-24}{25} $ ---(3).
We know that $ \sin 2x=\dfrac{2\tan x}{1+{{\tan }^{2}}x} $ . Let us use this results in equation (3).
 $ \Rightarrow \dfrac{2\tan x}{1+{{\tan }^{2}}x}=\dfrac{-24}{25} $ .
 $ \Rightarrow \dfrac{\tan x}{1+{{\tan }^{2}}x}=\dfrac{-12}{25} $ .
 $ \Rightarrow 25\tan x=-12-12{{\tan }^{2}}x $ .
 $ \Rightarrow 12{{\tan }^{2}}x+25\tan x+12=0 $ .
Let us factorize the obtained quadratic equation to find the value of $ \tan x $ .
 $ \Rightarrow 12{{\tan }^{2}}x+16\tan x+9\tan x+12=0 $ .
 $ \Rightarrow 4\tan x\left( 3\tan x+4 \right)+3\left( 3\tan x+4 \right)=0 $ .
 $ \Rightarrow \left( 4\tan x+3 \right)\left( 3\tan x+4 \right)=0 $ .
 $ \Rightarrow 4\tan x+3=0 $ , $ 3\tan x+4=0 $ .
 $ \Rightarrow 4\tan x=-3 $ , $ 3\tan x=-4 $ .
 $ \Rightarrow \tan x=\dfrac{-3}{4} $ , $ \tan x=\dfrac{-4}{3} $ .
∴ The value of $ \tan x $ is $ \dfrac{-3}{4} $ or $ \dfrac{-4}{5} $ .

Note:
We can see that the given problem consists of a huge amount of calculation so, we need to perform each step carefully. We can also solve this problem as shown below:
We have given $ \sin x+\cos x=\dfrac{1}{5} $ ---(4).
Let us square on both sides.
 $ \Rightarrow {{\left( \sin x+\cos x \right)}^{2}}={{\left( \dfrac{1}{5} \right)}^{2}} $ .
 $ \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x=\dfrac{1}{25} $ .
We know that $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ .
 $ \Rightarrow 1+2\sin x\cos x=\dfrac{1}{25} $ .
 $ \Rightarrow 2\sin x\cos x=\dfrac{-24}{25} $ .
 $ \Rightarrow -2\sin x\cos x=\dfrac{24}{25} $ .
 $ \Rightarrow 1-2\sin x\cos x=\dfrac{24}{25}+1 $ .
We know that $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ .
 $ \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x=\dfrac{24}{25}+1 $ .
 $ \Rightarrow {{\left( \sin x-\cos x \right)}^{2}}=\dfrac{49}{25} $ .
 $ \Rightarrow \sin x-\cos x=\pm \dfrac{7}{5} $ .
 $ \Rightarrow \sin x=\cos x\pm \dfrac{7}{5} $ ---(5).
Let us substitute equation (5) in equation (4).
So, we get $ \cos x\pm \dfrac{7}{5}+\cos x=\dfrac{1}{5} $ .
 $ \Rightarrow 2\cos x=\dfrac{1}{5}\pm \dfrac{7}{5} $ .
 $ \Rightarrow 2\cos x=\dfrac{8}{5}or\dfrac{-6}{5} $ .
 $ \Rightarrow \cos x=\dfrac{4}{5}or\dfrac{-3}{5} $ .
We know that $ \sec x=\dfrac{1}{\cos x} $ .
 $ \Rightarrow \sec x=\dfrac{5}{4}or\dfrac{-5}{3} $ .
 $ \Rightarrow {{\sec }^{2}}x=\dfrac{25}{16}or\dfrac{25}{9} $ .
 $ \Rightarrow {{\sec }^{2}}x-1=\dfrac{9}{16}or\dfrac{16}{9} $ .
We know that $ {{\sec }^{2}}x-1={{\tan }^{2}}x $ .
 $ \Rightarrow {{\tan }^{2}}x=\dfrac{9}{16}or\dfrac{16}{9} $ .
 $ \Rightarrow \tan x=\dfrac{-3}{4}or\dfrac{-4}{3} $ .