Question

If $\cos ec\theta -\sin \theta ={{a}^{3}}$ , $\sec \theta -\cos \theta ={{b}^{3}}$ then ${{a}^{2}}{{b}^{2}}({{a}^{2}}+{{b}^{2}})=$
(a) $2$
(b) ${{\sin }^{2}}\theta $
(c) ${{\cos }^{2}}\theta $
(d) $1$

Answer 1

Hint: For questions like these, we have to simplify the equations until they are in a form that can be added to give the simplest form of answer possible. Since cosec and secant are inverses of sine and cosine respectively, we will convert them into their respective inverses and then add them to get the answer.

Formula used: The identities used in these questions are the inverse of the basic trigonometric metrics. That is, cosec is the trigonometric inverse of sine function and similarly, secant is the trigonometric inverse of cosine function. This can be mathematically represented as;
$\cos ec\theta =\dfrac{1}{\sin \theta }$
And
$sec\theta =\dfrac{1}{\cos \theta }$
The identity to help simplify the equation will be the addition of squares of trigonometric functions
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
This can be rearranged accordingly to help get the required answer.

Complete step by step solution:
In the first given equation, we will convert cos to sine function and then simplify the function to find the value of $a$
$\cos ec\theta -\sin \theta ={{a}^{3}}$
$\Rightarrow \dfrac{1}{\sin \theta }-\sin \theta ={{a}^{3}}$
$\Rightarrow \dfrac{1-{{\sin }^{2}}\theta }{\sin \theta }={{a}^{3}}$
Rearranging the trigonometric identity of addition of squares
$\Rightarrow \dfrac{{{\cos }^{2}}\theta }{\sin \theta }={{a}^{3}}$
$\Rightarrow a={{(\dfrac{{{\cos }^{2}}\theta }{\sin \theta })}^{\dfrac{1}{3}}}$$...\left( 1 \right)$
Similarly in the second given equation, we will convert secant to cosine function and then simplify the function to find the value of $b$
$\sec \theta -\cos \theta ={{b}^{3}}$
$\Rightarrow \dfrac{1}{\cos \theta }-\cos \theta ={{b}^{3}}$
$\Rightarrow \dfrac{1-{{\cos }^{2}}\theta }{\cos \theta }={{b}^{3}}$
Rearranging the trigonometric identity of addition of squares
$\Rightarrow \dfrac{{{\sin }^{2}}\theta }{\cos \theta }={{b}^{3}}$
$\Rightarrow b={{(\dfrac{{{\sin }^{2}}\theta }{\cos \theta })}^{\dfrac{1}{3}}}$$...\left( 2 \right)$
The question is asking us to find the value of a particular expression. We will first simplify it and then plug in the values of $a$ and $b$ that we found by simplifying the trigonometric functions. This will give us the desired answer.
${{a}^{2}}{{b}^{2}}({{a}^{2}}+{{b}^{2}})={{a}^{4}}{{b}^{2}}+{{b}^{4}}{{a}^{2}}$
$\Rightarrow {{(\dfrac{{{\cos }^{2}}\theta }{\sin \theta })}^{\dfrac{4}{3}}}{{(\dfrac{{{\sin }^{2}}\theta }{\cos \theta })}^{\dfrac{2}{3}}}+{{(\dfrac{{{\cos }^{2}}\theta }{\sin \theta })}^{\dfrac{2}{3}}}{{(\dfrac{{{\sin }^{2}}\theta }{\cos \theta })}^{\dfrac{4}{3}}}$
$\Rightarrow {{(\dfrac{{{\cos }^{4}}\theta }{{{\sin }^{2}}\theta }\times \dfrac{{{\sin }^{2}}\theta }{\cos \theta })}^{\dfrac{2}{3}}}+{{(\dfrac{{{\cos }^{2}}\theta }{\sin \theta }\times \dfrac{{{\sin }^{4}}\theta }{{{\cos }^{2}}\theta })}^{\dfrac{2}{3}}}$
\[\Rightarrow {{(\dfrac{{{\cos }^{4}}\theta }{\cos \theta })}^{\dfrac{2}{3}}}+{{(\dfrac{{{\sin }^{4}}\theta }{\sin \theta })}^{\dfrac{2}{3}}}\]
\[\Rightarrow {{({{\cos }^{3}}\theta )}^{\dfrac{2}{3}}}+{{({{\sin }^{3}}\theta )}^{\dfrac{2}{3}}}\]
$\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ (This is an identity)

Therefore the answer is option (D)

Note: This question can also be solved by converting sine to cosec and cosine to secant but trigonometric identities wouldn’t really help there. The equations would be long and confusing. It is always more viable to convert to sine and cosine.
Here the rule of indices plays a big role because if you use it wrong then the answer will get wrong. There are many rules that come under indices; the rule which is used here is.
\[{{\left( {{\text{a}}^{x}} \right)}^{y}}=\text{ }{{\text{a}}^{x\times y}}\]
For example \[{{\left( {{\text{a}}^{3}} \right)}^{6}}={{\text{a}}^{3\times 6}}={{\text{a}}^{18}}\]