Question

Prove that $ (\sec A - \cos ecA)(1 + \tan A + \cot A) = \tan A\sec A - \cot A\cos ecA $

Answer 1

Hint: Simplify the trigonometric quantities in the form of sin and cos, then solve one side of one and using trigonometric functions and identities, try to make it equal to the other side. Also, rewrite the identities to check if they replace something in any equation.

Complete step-by-step answer:
We have to prove that $ (\sec A - \cos ecA)(1 + \tan A + \cot A) = \tan A\sec A - \cot A\cos ecA $
So, we start by solving Left Hand Side first:
We know that, $ \sec A = \dfrac{1}{{\cos A}} $ , $ \cos ecA = \dfrac{1}{{\sin A}} $ , $ \tan A = \dfrac{{\sin A}}{{\cos A}} $ and $ \cot A = \dfrac{{\cos A}}{{\sin A}} $ .
Putting the above values in the given equation, we get –
 $ (\dfrac{1}{{\cos A}} - \dfrac{1}{{\sin A}})(1 + \dfrac{{\sin A}}{{\cos A}} + \dfrac{{\cos A}}{{\sin A}}) $
On further solving, we get –
 $ (\dfrac{{\sin A - \cos A}}{{\sin A\cos A}})(\dfrac{{\sin A\cos A + {{\sin }^2}A + {{\cos }^2}A}}{{\sin A\cos A}}) $
We know that, $ {\sin ^2}A + {\cos ^2}A = 1 $ , putting this value in the above equation,
 $ (\dfrac{{\sin A - \cos A}}{{\sin A\cos A}})(\dfrac{{\sin A\cos A + 1}}{{\sin A\cos A}}) $
Multiplying the resultant values,
 $ \dfrac{{{{\sin }^2}A\cos A + \sin A - \sin A{{\cos }^2}A - \cos A}}{{{{\sin }^2}A{{\cos }^2}A}} $
Now $ {\sin ^2}A + {\cos ^2}A = 1 $ can also be written as,
 $ {\sin ^2}A = 1 - {\cos ^2}A $ and $ {\cos ^2}A = 1 - {\sin ^2}A $ .
Putting the above two results in the simplified equation, we get
 $
\Rightarrow \dfrac{{(1 - {{\cos }^2}A)\cos A + \sin A - \sin A(1 - {{\sin }^2}A) - \cos A}}{{{{\sin }^2}A{{\cos }^2}A}} \\
   = \dfrac{{\cos A - {{\cos }^3}A + \sin A - \sin A + {{\sin }^3}A - cosA}}{{{{\sin }^2}A{{\cos }^2}A}} \\
   = \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{{{\sin }^2}A{{\cos }^2}A}} \\
   = \dfrac{{{{\sin }^3}A}}{{{{\sin }^2}A{{\cos }^2}A}} - \dfrac{{{{\cos }^3}A}}{{{{\sin }^2}A{{\cos }^2}A}} \\
   = \dfrac{{\sin A}}{{{{\cos }^2}A}} - \dfrac{{\cos A}}{{{{\sin }^2}A}} \\
   = \tan A\sec A - \cot A\cos ecA \;
  $
Now Left Hand Side = Right Hand Side
Hence proved.

Note: There are six trigonometric ratios, sine, cosine, tangent, cotangent, cosecant and secant. These six trigonometric ratios are abbreviated as sin, cos, tan, cot, cosec and sec respectively. Since they are expressed in terms of the ratio of sides of a right angled triangle for a specific angle θ , they are called trigonometric ratios.