Question

How do you express ${\cot ^3}\theta - {\cos ^2}\theta - {\tan ^3}\theta$ in terms of non-exponential trigonometric function?

Answer 1

To solve this question first we use the algebraic identity and convert the equation in sin and cos trigonometric function and then take LCM in the denominator and use the identity and formulas and try to convert the whole relations in the power one and that is the final answer.

Complete step-by-step answer:
Given,
${\cot ^3}\theta - {\cos ^2}\theta - {\tan ^3}\theta$
We have to express this term in non-exponential trigonometric functions.
First we are using the algebraic identity ${a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$
We have to use this identity $\cot \theta$ and $\tan \theta$
$\Rightarrow {\cot ^3}\theta - {\cos ^2}\theta - {\tan ^3}\theta$
Using the identity of ${a^3} - {b^3}$
$\Rightarrow \left( {\cot \theta - \tan \theta } \right)\left( {{{\cot }^2}\theta + {{\tan }^2}\theta + \cot \theta \tan \theta } \right) - {\cos ^2}\theta$
We know that multiplication of $\cot \theta$ and $\tan \theta$ is 1.
Now we are converting all the functions in sin and cos function.
$\Rightarrow \left( {\dfrac{{\cos \theta }}{{\sin \theta }} - \dfrac{{\sin \theta }}{{\cos \theta }}} \right)\left( {\dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} + \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} + 1} \right) - {\cos ^2}\theta$
Now taking LCM in denominator
$\Rightarrow \left( {\dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{\sin \theta \cos \theta }}} \right)\left( {\dfrac{{{{\cos }^4}\theta + {{\sin }^4}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }} + 1} \right) - {\cos ^2}\theta$
Converting first multiplication in double angle by multiplying and dividing by 2
On using the formula of trigonometry ${\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta$ and $2\sin \theta \cos \theta = \sin 2\theta$
$\Rightarrow \left( {\dfrac{{2\cos 2\theta }}{{\sin 2\theta }}} \right)\left( {\dfrac{{{{\cos }^4}\theta + {{\sin }^4}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }} + 1} \right) - {\cos ^2}\theta$
Now taking the LCM in the second multiplication and adding and subtracting the same part one time
$\Rightarrow \left( {\dfrac{{2\cos 2\theta }}{{\sin 2\theta }}} \right)\left( {\dfrac{{{{\cos }^4}\theta + {{\sin }^4}\theta + 2{{\sin }^2}\theta {{\cos }^2}\theta - {{\sin }^2}\theta {{\cos }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} \right) - {\cos ^2}\theta$
Now making the perfect square in second multiple.
$\Rightarrow \left( {\dfrac{{2\cos 2\theta }}{{\sin 2\theta }}} \right)\left( {\dfrac{{{{\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)}^2} - {{\sin }^2}\theta {{\cos }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} \right) - {\cos ^2}\theta$
Now using the identity of trigonometry ${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\Rightarrow \left( {\dfrac{{2\cos 2\theta }}{{\sin 2\theta }}} \right)\left( {\dfrac{{1 - {{\sin }^2}\theta {{\cos }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} \right) - {\cos ^2}\theta$
Now using the formulas $\dfrac{{\cos 2\theta }}{{\sin 2\theta }} = \cot 2\theta$, ${\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2}$ and splitting the second multiplication.
$\Rightarrow \left( {2\cot 2\theta } \right)\left( {\dfrac{1}{{{{\sin }^2}\theta {{\cos }^2}\theta }} - 1} \right) - \dfrac{{1 + \cos 2\theta }}{2}$
Multiplying and dividing by 4 in the second multiplication
$\Rightarrow \left( {2\cot 2\theta } \right)\left( {\dfrac{4}{{{{\sin }^2}2\theta }} - 1} \right) - \dfrac{{1 + \cos 2\theta }}{2}$
Now using the formula ${\sin ^2}2\theta = \dfrac{{1 - \cos 4\theta }}{2}$
$\Rightarrow \left( {2\cot 2\theta } \right)\left( {\dfrac{8}{{1 - \cos 2\theta }} - 1} \right) - \dfrac{{1 + \cos 2\theta }}{2}$
Final answer:
$\Rightarrow \left( {2\cot 2\theta } \right)\left( {\dfrac{8}{{1 - \cos 2\theta }} - 1} \right) - \dfrac{{1 + \cos 2\theta }}{2}$ is the non-exponential expression of ${\cot ^3}\theta - {\cos ^2}\theta - {\tan ^3}\theta$

Note: This question is tricky as well as difficult you must solve this type of question. To solve these types of questions we must know all the formulas, identities, and the relation between all the trigonometric relations. There are many places where students often make mistakes to take a look at clearly.