Question

The expression ${\operatorname{cosec} ^2}A{\cot ^2}A - {\sec ^2}A{\tan ^2}A - \left( {{{\cot }^2}A - {{\tan }^2}A} \right)\left( {{{\sec }^2}A + {{\operatorname{cosec} }^2}A - 1} \right)$ is equal to
A) $0$
B) $1$
C) ${\sec ^2}A$
D) ${\operatorname{cosec} ^2}A$

Answer 1

Hint: In this question, use the trigonometry properties of $\sin x$ and $\cos x$ for solving these types of questions. Also use the algebraic formula of cubic and squares the terms for expanding the term to simplify the expression.

Complete step by step solution:
First, we write required trigonometric properties as shown below,
$ \Rightarrow {\sin ^2}x + {\cos ^2}x = 1$
$ \Rightarrow \tan x = \dfrac{{\sin x}}{{\cos x}}$
$ \Rightarrow \sec x = \dfrac{1}{{\cos x}}$
Now, we write the required algebraic formulas as,
$ \Rightarrow {x^3} - {y^3} = \left( {x - y} \right)\left( {{x^2} + {y^2} + xy} \right)$
$ \Rightarrow {x^2} + {y^2} = {\left( {x + y} \right)^2} - 2xy$
In this question, we have given trigonometric expression as,
${\operatorname{cosec} ^2}A{\cot ^2}A - {\sec ^2}A{\tan ^2}A - \left( {{{\cot }^2}A - {{\tan }^2}A} \right)\left( {{{\sec }^2}A + {{\operatorname{cosec} }^2}A - 1} \right)$
Now we will break the given trigonometric expression into two parts or the trigonometric expression can be written as \[P - Q\].
Here, $P = {\operatorname{cosec} ^2}A{\cot ^2}A - {\sec ^2}A{\tan ^2}A$ and $Q = \left( {{{\cot }^2}A - {{\tan }^2}A} \right)\left( {{{\sec }^2}A + {{\operatorname{cosec} }^2}A - 1} \right)$
Now we can write the expression as,$ \Rightarrow {\operatorname{cosec} ^2}A{\cot ^2}A - {\sec ^2}A{\tan ^2}A - \left( {{{\cot }^2}A - {{\tan }^2}A} \right)\left( {{{\sec }^2}A + {{\operatorname{cosec} }^2}A - 1} \right) = P - Q$
First, we will solve the term $P$ by expanding the $\csc $ and $\cot $ terms in $\sin $ and $\cos $ terms because it is easy to solve and calculate.
$ \Rightarrow P = {\operatorname{cosec} ^2}A{\cot ^2}A - {\sec ^2}A{\tan ^2}A$
Now we will apply the trigonometric properties as,
$ \Rightarrow P = \dfrac{1}{{{{\sin }^2}A}} \times \dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}} - \dfrac{1}{{{{\cos }^2}A}} \times \dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}$
Now, we will simplify the above expression as,
$ \Rightarrow P = \dfrac{{{{\cos }^2}A}}{{{{\sin }^4}A}} - \dfrac{{{{\sin }^2}A}}{{{{\cos }^4}A}}$
After simplification we will get,
$ \Rightarrow P = \dfrac{{{{\cos }^6}A - {{\sin }^6}A}}{{\left( {{{\sin }^4}A} \right)\left( {{{\cos }^4}A} \right)}}$
Now we will use the algebraic formula in the numerator of the expression to find the easiest form and then solve it accordingly.
$ \Rightarrow P = \dfrac{{{{\left( {{{\cos }^2}A} \right)}^3} - {{\left( {{{\sin }^2}A} \right)}^3}}}{{\left( {{{\sin }^4}A} \right)\left( {{{\cos }^4}A} \right)}}$
Now, we will apply the algebraic properties as,
$ \Rightarrow P = \dfrac{{\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\left( {{{\cos }^4}A + {{\sin }^4}A + {{\sin }^2}A{{\cos }^2}A} \right)}}{{\left( {{{\sin }^4}A} \right)\left( {{{\cos }^4}A} \right)}}$
Now, we will simplify the above expression as,
$ \Rightarrow P = \dfrac{{\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\left( {{{\left( {{{\cos }^2}A + {{\sin }^2}A} \right)}^2} - {{\sin }^2}A{{\cos }^2}A} \right)}}{{\left( {{{\sin }^4}A} \right)\left( {{{\cos }^4}A} \right)}}$
After simplification we will get,
$ \Rightarrow P = \dfrac{{\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\left( {1 - {{\sin }^2}A{{\cos }^2}A} \right)}}{{\left( {{{\sin }^4}A} \right)\left( {{{\cos }^4}A} \right)}}$
Now we will solve the term $Q$ by expanding the $\csc $ and $\cot $ terms in $\sin $ and $\cos $terms because it is easy to solve and calculate.
$ \Rightarrow Q = \left( {\dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}} - \dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}} \right)\left( {\dfrac{1}{{{{\cos }^2}A}} + \dfrac{1}{{{{\sin }^2}A}} - 1} \right)$
Now, we will apply the algebraic properties as,
$ \Rightarrow Q = \left( {\dfrac{{{{\cos }^4}A - {{\sin }^4}A}}{{\left( {{{\sin }^2}A} \right)\left( {{{\cos }^2}A} \right)}}} \right)\left( {\dfrac{{{{\sin }^2}A + {{\cos }^2}A - {{\sin }^2}A{{\cos }^2}A}}{{{{\sin }^2}A{{\cos }^2}A}}} \right)$
Now, we will simplify the above expression as,
$ \Rightarrow Q = \left( {\dfrac{{{{\cos }^4}A - {{\sin }^4}A}}{{\left( {{{\sin }^2}A} \right)\left( {{{\cos }^2}A} \right)}}} \right)\left( {\dfrac{{1 - {{\sin }^2}A{{\cos }^2}A}}{{{{\sin }^2}A{{\cos }^2}A}}} \right)$
After simplification we will get,
$ \Rightarrow Q = \dfrac{{\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\left( {1 - {{\sin }^2}A{{\cos }^2}A} \right)}}{{\left( {{{\sin }^4}A} \right)\left( {{{\cos }^4}A} \right)}}$
Now, we will substitute $\dfrac{{\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\left( {1 - {{\sin }^2}A{{\cos }^2}A} \right)}}{{\left( {{{\sin }^4}A} \right)\left( {{{\cos }^4}A} \right)}}$ for $P$ and $\dfrac{{\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\left( {1 - {{\sin }^2}A{{\cos }^2}A} \right)}}{{\left( {{{\sin }^4}A} \right)\left( {{{\cos }^4}A} \right)}}$ for$Q$ in the expression \[P - Q\] to find the solution of the trigonometric expression.
\[ \Rightarrow P - Q = \left[
  \dfrac{{\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\left( {1 - {{\sin }^2}A{{\cos }^2}A} \right)}}{{\left( {{{\sin }^4}A} \right)\left( {{{\cos }^4}A} \right)}} \\
   - \dfrac{{\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\left( {1 - {{\sin }^2}A{{\cos }^2}A} \right)}}{{\left( {{{\sin }^4}A} \right)\left( {{{\cos }^4}A} \right)}} \\ \right]\]
After simplification, we will get
\[\therefore P - Q = 0\]
Therefore, the solution of the trigonometric expression ${\operatorname{cosec} ^2}A{\cot ^2}A - {\sec ^2}A{\tan ^2}A - \left( {{{\cot }^2}A - {{\tan }^2}A} \right)\left( {{{\sec }^2}A + {{\operatorname{cosec} }^2}A - 1} \right)$ is $0$.

Note: As we know that trigonometry is the most important and useful topic of mathematics. There is some basic formula used to solve the questions or convert the easiest form. The trigonometric expression represents the relation between the sine and cosine and tangent of the function, or it can be said that it is a reciprocal relation of sine and cosine.