Question

Prove that $\left( {1 + \cot a - \cos eca} \right)\left( {1 + \tan a - \sec a} \right) = 2$.

Answer 1

Hint : First, we need to analyze the given information so that we can able to solve the problem. Generally, in Mathematics, the trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Here we are asked to prove that $\left( {1 + \cot a - \cos eca} \right)\left( {1 + \tan a - \sec a} \right) = 2$
We need to apply the appropriate trigonometric identities to obtain the required answer.

Formula used:
The trigonometric identities that are used to solve the given problem are as follows.
a) $\tan x = \dfrac{{\sin x}}{{\cos x}}$
b) $\cot x = \dfrac{{\cos x}}{{\sin x}}$
c) $\sec x = \dfrac{1}{{\cos x}}$
d) $\cos ecx = \dfrac{1}{{\sin x}}$
e)$\left( {x + y} \right)\left( {x - y} \right) = {x^2} - {y^2}$
f) ${\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$
g) ${\sin ^2}x + {\cos ^2}x = 1$

Complete step-by-step solution:
Here, we are asked to prove that $\left( {1 + \cot a - \cos eca} \right)\left( {1 + \tan a - \sec a} \right) = 2$
Let us consider the left-hand side of the equation and we need to show that while solving the left-hand side of the equation, we get the answer $2$
We need to apply the appropriate trigonometric identities to obtain the required answer.

$LHS = \left( {1 + \cot a - \cos eca} \right)\left( {1 + \tan a - \sec a} \right)$
Now, we shall apply the following trigonometric identities $\tan x = \dfrac{{\sin x}}{{\cos x}}$ , $\cot x = \dfrac{{\cos x}}{{\sin x}}$ , $\sec x = \dfrac{1}{{\cos x}}$ and $\cos ecx = \dfrac{1}{{\sin x}}$ on the above.
We get, $LHS = \left( {1 + \dfrac{{\cos a}}{{\sin a}} - \dfrac{1}{{\sin a}}} \right)\left( {1 + \dfrac{{\sin a}}{{\cos a}} - \dfrac{1}{{\cos a}}} \right)$
Taking LCM inside the brackets, we have
$LHS = \left( {\dfrac{{\sin a + \cos a - 1}}{{\sin a}}} \right)\left( {\dfrac{{\cos a + \sin a - 1}}{{\cos a}}} \right)$
$= \dfrac{{\left( {\sin a + \cos a - 1} \right)\left( {\cos a + \sin a - 1} \right)}}{{\sin a\cos a}}$
Since $x = \sin a + \cos b$ and $y = 1$ , we can apply the formula $\left( {x + y} \right)\left( {x - y} \right) = {x^2} - {y^2}$ here.
Thus $LHS = \dfrac{{{{\left( {\sin a + \cos a} \right)}^2} - {1^2}}}{{\sin a\cos a}}$
Now, applying the formula ${\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$ , we have,
$LHS = \dfrac{{{{\sin }^2}a + {{\cos }^2}a + 2\sin a\cos a - 1}}{{\sin a\cos a}}$
Since we know that ${\sin ^2}x + {\cos ^2}x = 1$ , we get
$LHS = \dfrac{{1 + 2\sin a\cos a - 1}}{{\sin a\cos a}}$
$= \dfrac{{2\sin a\cos a}}{{\sin a\cos a}}$
$= 2$ which we are asked to prove.
Hence we showed that $\left( {1 + \cot a - \cos eca} \right)\left( {1 + \tan a - \sec a} \right) = 2$

Note: Generally, trigonometric identities are equalities that involve trigonometric functions and are useful whenever trigonometric functions are involved in an expression or an equation. The six basic trigonometric ratios are sine, cosine, tangent, cosecant, secant, and cotangent and all the fundamental trigonometric identities are derived from the six trigonometric ratios. Hence we showed that $\left( {1 + \cot a - \cos eca} \right)\left( {1 + \tan a - \sec a} \right) = 2$