Question

How do you solve ${\left( {1 + \tan A} \right)^2} + {\left( {1 + \cot A} \right)^2} = {\left( {\sec A + \cos ecA} \right)^2}$?

Answer 1

Here, in the given question, we need to solve ${\left( {1 + \tan A} \right)^2} + {\left( {1 + \cot A} \right)^2} = {\left( {\sec A + \cos ecA} \right)^2}$, and find the values of $A$. At first, we will convert all the trigonometric functions in term of $\sin$ and $\cos$, after this we will take LCM on both sides of the given equation and simplify the equation and cancel-out the common factors. At the end we will find the values of $A$.

Complete step by step answer:
We have, ${\left( {1 + \tan A} \right)^2} + {\left( {1 + \cot A} \right)^2} = {\left( {\sec A + \cos ecA} \right)^2}$. Let us convert all the trigonometric functions in the form of $\sin$ and $\cos$.As we know $\tan x = \dfrac{{\sin x}}{{\cos x}}$, $\cot x = \dfrac{{\cos x}}{{\sin x}}$, $\sec x = \dfrac{1}{{\cos x}}$ and $\cos ecx = \dfrac{1}{{\sin x}}$. Therefore, we get
$\Rightarrow {\left( {1 + \dfrac{{\sin A}}{{\cos A}}} \right)^2} + {\left( {1 + \dfrac{{\cos A}}{{\sin A}}} \right)^2} = {\left( {\dfrac{1}{{\cos A}} + \dfrac{1}{{\sin A}}} \right)^2}$

Take LCM on both sides.
$\Rightarrow \dfrac{{{{\left( {\cos A + \sin A} \right)}^2}}}{{{{\cos }^2}A}} + \dfrac{{{{\left( {\sin A + \cos A} \right)}^2}}}{{{{\sin }^2}A}} = \dfrac{{{{\left( {\sin A + \cos A} \right)}^2}}}{{\sin A\cos A}}$
Take ${\left( {\cos A + \sin A} \right)^2}$ as a common term.
$\Rightarrow {\left( {\cos A + \sin A} \right)^2}\left( {\dfrac{1}{{{{\cos }^2}A}} + \dfrac{1}{{{{\sin }^2}A}}} \right) = \dfrac{{{{\left( {\sin A + \cos A} \right)}^2}}}{{\sin A\cos A}}$
On canceling-out common terms on both sides, we get
$\Rightarrow \dfrac{1}{{{{\cos }^2}A}} + \dfrac{1}{{{{\sin }^2}A}} = \dfrac{1}{{\sin A\cos A}}$
Take LCM on the left-hand side
$\Rightarrow \dfrac{{{{\sin }^2}A + {{\cos }^2}A}}{{{{\sin }^2}A \times {{\cos }^2}A}} = \dfrac{1}{{\sin A\cos A}}$

As we know ${\sin ^2}A + {\cos ^2}A = 1$. Therefore, we get
$\Rightarrow \dfrac{1}{{{{\left( {\sin A\cos A} \right)}^2}}} = \dfrac{1}{{\sin A\cos A}}$
On reciprocating both sides, we get
$\Rightarrow {\left( {\sin A\cos A} \right)^2} = \sin A\cos A$
$\Rightarrow {\left( {\sin A\cos A} \right)^2} - \sin A\cos A = 0$
Take $\sin A\cos A$ as a common factor.
$\Rightarrow \left( {\sin A\cos A} \right)\left( {\sin A\cos A - 1} \right) = 0$
Now, we will find the values $A$.
$\Rightarrow \sin A\cos A = 0$
From here we get,
$\Rightarrow \sin A = 0$ and $\cos A = 0$

For $\sin A = 0$, we have
$A = 0,\pi ,2\pi ,3\pi ,....$
For $\cos A = 0$, we have
$A = \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2},\dfrac{{7\pi }}{2},.....$
$\Rightarrow \sin A\cos A = 0$
As we know, $\sin 2A = 2\sin A\cos A$, from here we get $\sin A\cos A = \dfrac{1}{2}\sin 2A$.
$\Rightarrow \dfrac{1}{2}\sin 2A = 0$
$\Rightarrow \sin 2A = 0$
From here, we get
$2A = 0,\pi ,2\pi ,3\pi ,.....$
On dividing by $2$, we get
$A = 0,\dfrac{\pi }{2},\pi ,\dfrac{{3\pi }}{2},2\pi ,\dfrac{{5\pi }}{2},3\pi ,\dfrac{{7\pi }}{2},.....$

Hence, the values of $A$ satisfying the equation ${\left( {1 + \tan A} \right)^2} + {\left( {1 + \cot A} \right)^2} = {\left( {\sec A + \cos ecA} \right)^2}$ are, $A = ....,\dfrac{{ - 7\pi }}{2}, - 3\pi ,\dfrac{{ - 5\pi }}{2}, - 2\pi ,\dfrac{{ - 3\pi }}{2}, - \pi ,\dfrac{{ - \pi }}{2},0,\dfrac{\pi }{2},\pi ,\dfrac{{3\pi }}{2},2\pi ,3\pi ,\dfrac{{7\pi }}{2},...$

Note: To solve this type of question try to convert the different functions in terms of two or three functions only and simplify it using identities. One must know all the trigonometric formulas to solve these types of questions and one must remember all the trigonometric values. To solve these type of questions we should know all the required values of standard angles say, $0^\circ ,30^\circ ,60^\circ ,90^\circ ,180^\circ ,270^\circ ,360^\circ$ respectively for each trigonometric term such as $\sin ,\cos ,\tan ,\cos ec,\sec ,\cot$.