Question

How do you verify $ \sin x\left( {\tan x\cos x - \cot x\cos x} \right) = 1 - 2{\cos ^2}x $ ?

Answer 1

Hint: Here, you are given an identity which consists of the trigonometric ratio sine, tangent, cosine and cotangent and are asked to prove the identity. What you need to do here is use the trigonometric identities such as $ \tan x = \dfrac{{\sin x}}{{\cos x}} $ , $ \cot x = \dfrac{1}{{\tan x}} $ and $ {\sin ^2}x + {\cos ^2}x = 1 $ and simplify the equation. By simplifying what is meant is that bring down the left-hand side equal to the right-hand side. Also, as you can see that the right-hand side consists of only cosine, it is better that you convert all the ratio present on the left-hand side into sine and cosine.

Complete step by step solution:
Let us consider the left-hand side of the identity given to you and convert all ratios into sine and cosine. You know that tangent of any angle is equal to the sine of that angle divided by cosine of that angle, mathematically, we have $ \tan x = \dfrac{{\sin x}}{{\cos x}} $ and also the cotangent is equal to the inverse of tangent, so you can write cotangent of that angle as $ \cot x = \dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}} $ . Let us substitute these values of $ \tan x $ and $ \cot x $ in the left-hand side expression. We get,
 $
  \sin x\left( {\tan x\cos x - \cot x\cos x} \right) \\
   = \sin x\left( {\dfrac{{\sin x}}{{\cos x}}\cos x - \dfrac{{\cos x}}{{\sin x}}\cos x} \right) \\
   = \sin x\left( {\sin x - \dfrac{{{{\cos }^2}x}}{{\sin x}}} \right) \\
  $
In order to subtract $ \dfrac{{{{\cos }^2}x}}{{\sin x}} $ from $ \sin x $ , the denominator of the terms should be equal and therefore we multiply and divide $ \sin x $ by $ \sin x $ , we get,
 $ \sin x\left( {\sin x - \dfrac{{{{\cos }^2}x}}{{\sin x}}} \right) = \sin x\left( {\dfrac{{{{\sin }^2}x}}{{\sin x}} - \dfrac{{{{\cos }^2}x}}{{\sin x}}} \right) = \sin x\left( {\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x}}} \right) = {\sin ^2}x - {\cos ^2}x $
Now, use the identity which gives you the sum of squares of $ \sin x $ and $ \cos x $ is equal to one, mathematically, $ {\sin ^2}x + {\cos ^2}x = 1 $ . From here we can write $ {\sin ^2}x = 1 - {\cos ^2}x $ .Substitute this value of $ {\sin ^2}x $ in the above obtained equation, we get,
 $
  \sin x\left( {\tan x\cos x - \cot x\cos x} \right) = \left( {1 - {{\cos }^2}x} \right) - {\cos ^2}x \\
  \sin x\left( {\tan x\cos x - \cot x\cos x} \right) = 1 - 2{\cos ^2}x \\
  $
Hence proved $ \sin x\left( {\tan x\cos x - \cot x\cos x} \right) = 1 - 2{\cos ^2}x $ , identity verified.

Note: Here, the main idea to apply was to convert all the trigonometric ratio to sine and cosine. This idea should strike by looking at the right-hand side of the equation as it contains only cosine and hence it indicates that the left-hand side should contain sine and cosine so that further we can write cosine in terms of sine. You should memorize all the basic and fundamental properties such as $ \tan x = \dfrac{{\sin x}}{{\cos x}} $ , $ \cot x = \dfrac{1}{{\tan x}} $ and $ {\sin ^2}x + {\cos ^2}x = 1 $ .