Question

If \[{\rm{cosec}}\theta - \sin \theta = {a^3}\]; \[\sec \theta - \cos \theta = {b^3}\] then, \[{a^2}{b^2}\left[ {{a^2} + {b^2}} \right] = \]
A) 1
B) 2
C) \[{\sin ^2}\theta \]

Answer 1

Hint:
We are required to find the value of the expression given to us. The values of the variables are given to us in the form of trigonometric expressions. We will simplify these trigonometric expressions using the various trigonometric formulas and identities. We will then substitute these simplified trigonometric expressions in the given expression and manipulate it further to find its value.
Formula Used: We will use the following trigonometric formulas and identities to solve our question –
\[\begin{array}{l}{\rm{cosec}}\theta = \dfrac{1}{{\sin \theta }}\\\sec \theta = \dfrac{1}{{\cos \theta }}\\{\sin ^2}\theta + {\cos ^2}\theta = 1\end{array}\]

Complete step by step solution:
We are given that \[{\rm{cosec}}\theta - \sin \theta = {a^3}\]
We will first solve and simplify the LHS of this equation.
We can write \[{\rm{cosec}}\theta \] in terms of \[\sin \theta \]using the formula \[{\rm{cosec}}\theta = \dfrac{1}{{\sin \theta }}\]. Taking the LCM of the denominator and solving further, the LHS becomes,
\[\begin{array}{c}{\rm{cosec}}\theta - \sin \theta = \dfrac{1}{{\sin \theta }} - \sin \theta \\ = \dfrac{1}{{\sin \theta }} - \dfrac{{\sin \theta \left[ {\sin \theta } \right]}}{{\sin \theta }}\end{array}\]
\[{\rm{cosec}}\theta - \sin \theta = \dfrac{{1 - {{\sin }^2}\theta }}{{\sin \theta }}\]…………\[\left[ 1 \right]\]
Now we know \[{\sin ^2}\theta + {\cos ^2}\theta = 1\].
Hence, \[1 - {\sin ^2}\theta = {\cos ^2}\theta \]. Substitute this in equation \[\left[ 1 \right]\], we get,
\[{\rm{cosec}}\theta - \sin \theta = \dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}\]
Now we will equate the LHS to RHS, and find the value of \[a\].
\[\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }} = {a^3}\]
\[ \Rightarrow a = {\left[ {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right]^{\dfrac{1}{3}}}\]……………….\[\left[ 2 \right]\]
We are also given that \[{\rm{sec}}\theta - \cos \theta = {b^3}\]
We will now solve and simplify the LHS of this equation.
We can write \[{\rm{sec}}\theta \] in terms of \[\cos \theta \] using the formula \[{\rm{sec}}\theta = \dfrac{1}{{\cos \theta }}\]. Taking the LCM of the denominator and solving further, the LHS becomes,
\[\begin{array}{c}{\rm{sec}}\theta - \cos \theta = \dfrac{1}{{\cos \theta }} - \cos \theta \\ = \dfrac{1}{{\cos \theta }} - \dfrac{{\cos \theta \left[ {\cos \theta } \right]}}{{\cos \theta }}\end{array}\]
\[{\rm{sec}}\theta - \cos \theta = \dfrac{{1 - {{\cos }^2}\theta }}{{\cos \theta }}\]………….\[\left[ 3 \right]\]
Now we know \[{\sin ^2}\theta + {\cos ^2}\theta = 1\].
Hence, \[1 - {\cos ^2}\theta = {\sin ^2}\theta \]. Substituting this in equation \[\left[ 3 \right]\], we get,
\[\dfrac{{1 - {{\cos }^2}\theta }}{{\cos \theta }} = \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}\]
Now equate the LHS to RHS, and find the value of \[b\].
\[\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }} = {b^3}\]

\[b = {\left[ {\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right]^{\dfrac{1}{3}}}\]………………..\[\left[ 4 \right]\]
Now we will consider the expression \[{a^2}{b^2}\left[ {{a^2} + {b^2}} \right]\]. Multiplying the terms, we get
\[{a^2}{b^2}\left[ {{a^2} + {b^2}} \right] = {a^4}{b^2} + {a^2}{b^4}\]…………\[\left[ 5 \right]\]
Now we will substitute the values of \[a\]and \[b\]from equation \[\left[ 2 \right]\] and equation \[\left[ 4 \right]\] respectively in equation \[\left[ 5 \right]\]. We will then multiply the powers further to simplify the equation. On doing so, we get,
\[\begin{array}{c}{a^4}{b^2} + {a^2}{b^4} = {\left\{ {{{\left[ {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right]}^{\dfrac{1}{3}}}} \right\}^4}{\left\{ {{{\left[ {\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right]}^{\dfrac{1}{3}}}} \right\}^2} + {\left\{ {{{\left[ {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right]}^{\dfrac{1}{3}}}} \right\}^2}{\left\{ {{{\left[ {\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right]}^{\dfrac{1}{3}}}} \right\}^4}\\ = {\left[ {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right]^{\dfrac{4}{3}}}{\left[ {\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right]^{\dfrac{2}{3}}} + {\left[ {\dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right]^{\dfrac{2}{3}}}{\left[ {\dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right]^{\dfrac{4}{3}}}\end{array}\]
Now let us compare the powers of the various terms.
\[{a^4}{b^2} + {a^2}{b^4} = {\left[ {\cos \theta } \right]^{2 \times \dfrac{4}{3} - \dfrac{2}{3}}}{\left[ {\sin \theta } \right]^{2 \times \dfrac{2}{3} - \dfrac{4}{3}}} + {\left[ {\sin \theta } \right]^{2 \times \dfrac{4}{3} - \dfrac{2}{3}}}{\left[ {\cos \theta } \right]^{2 \times \dfrac{2}{3} - \dfrac{4}{3}}}\]
We will now subtract the powers,
\[\begin{array}{c}{a^4}{b^2} + {a^2}{b^4} = {\left[ {\cos \theta } \right]^{\dfrac{8}{3} - \dfrac{2}{3}}}{\left[ {\sin \theta } \right]^{\dfrac{4}{3} - \dfrac{4}{3}}} + {\left[ {\sin \theta } \right]^{\dfrac{8}{3} - \dfrac{2}{3}}}{\left[ {\cos \theta } \right]^{\dfrac{4}{3} - \dfrac{4}{3}}}\\ = {\left[ {\cos \theta } \right]^{\dfrac{6}{3}}}{\left[ {\sin \theta } \right]^0} + {\left[ {\sin \theta } \right]^{\dfrac{6}{3}}}{\left[ {\cos \theta } \right]^0}\end{array}\]
On simplifying the powers further, we get,
\[\begin{array}{c}{a^4}{b^2} + {a^2}{b^4} = {\left[ {\cos \theta } \right]^2} + {\left[ {\sin \theta } \right]^2}\\ = {\cos ^2}\theta + {\sin ^2}\theta \end{array}\]
Now we know \[{\sin ^2}\theta + {\cos ^2}\theta = 1\].
Substituting \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] in the equation, we get
\[{a^4}{b^2} + {a^2}{b^4} = 1\]

Hence, the value of the expression \[{a^2}{b^2}\left[ {{a^2} + {b^2}} \right]\] is 1.
Since, this answer matches with the option [A]; hence, option [A] is the correct answer.

Note:
Here, we need to remember all the trigonometric identities and formulas to solve the question. There are three primary functions of trigonometry; they are sine, tangent and cosine. Other three functions [cosecant, secant and cotangent] can be easily derived from the primary functions of trigonometry. Trigonometry is used in our real life for calculating height and distance also it is used in sectors like the department of marine, criminology etc.