Question

How do you prove $\left( {\sin x + \cos x} \right)\left( {\tan x + \cot x} \right) = \sec x + \csc x$?

Answer 1

Hint:In order to proof the above statement ,take the left hand side of the equation and put$\tan x = \dfrac{{\sin x}}{{\cos x}},\cot x = \dfrac{{\cos x}}{{\sin x}}$.now taking LCM and combining terms ,you will get \[{\sin ^2}x + {\cos ^2}x\] in the numerator put it equal to 1 according to the identity ,then separating the denominator further will give your final result which is equal to right-hand side of the equation.

Complete step by step solution:
To prove: $\left( {\sin x + \cos x} \right)\left( {\tan x + \cot x} \right) = \sec x + \csc x$

Proof:
Taking Left-hand Side of the equation,
$ \Rightarrow \left( {\sin x + \cos x} \right)\left( {\tan x + \cot x} \right)$

As we know that $\tan x$is equal to the ratio of $\sin x$to $\cos x$and similarly $\cot x$is equal to the ratio of $\cos x$to $\sin x$.In simple words,$\tan x = \dfrac{{\sin x}}{{\cos x}},\cot x = \dfrac{{\cos x}}{{\sin x}}$

Putting these values in the above equation, we get
 \[ \Rightarrow \left( {\sin x + \cos x} \right)\left( {\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin
x}}} \right)\]

Simplifying further by taking LCM as \[\cos x\sin x\] , we get
 \[ \Rightarrow \left( {\sin x + \cos x} \right)\left( {\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\cos x\sin
x}}} \right)\]

From the identity of trigonometry which states that sum of squares of sine and cosine is equal to 1
i.e. \[{\sin ^2}x + {\cos ^2}x = 1\] .

Replacing in the equation, we get
 \[
\Rightarrow \left( {\sin x + \cos x} \right)\left( {\dfrac{1}{{\cos x\sin x}}} \right) \\
\Rightarrow \dfrac{{\left( {\sin x + \cos x} \right)}}{{\cos x\sin x}} \\
\]
Separating the denominator
 \[
\Rightarrow \dfrac{{\sin x}}{{\cos x\sin x}} + \dfrac{{\cos x}}{{\cos x\sin x}} \\
\Rightarrow \dfrac{1}{{\cos x}} + \dfrac{1}{{\sin x}} \\
\]
As we know that \[\dfrac{1}{{\cos x}} = \sec x\] and \[\dfrac{1}{{\sin x}} = \csc x\]

 \[LHS = \sec x + \csc x\]

Taking Right-hand Side part of the equation
$RHS = \sec x + \csc x$
$\therefore LHS = RHS$

Hence, proved.

Additional Information:
1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.

2. Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$for all x in its domain.

Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$for all x in its domain.

We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta $

Note: 1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.