Prove the following identities $\dfrac{{{{\tan }^3}x}}{{1 + {{\tan }^2}x}} + \dfrac{{{{\cot }^3}x}}{{1 + {{\cot }^2}x}} = \dfrac{{1 - 2{{\sin }^2}x{{\cos }^2}x}}{{\sin x\cos x}}$
Answer
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Hint: To prove this question we must know following trigonometric identities which are mentioned below: -
$1 + {\tan ^2}\theta = {\sec ^2}\theta $
$1 + {\cot ^2}\theta = \cos e{c^2}\theta $
$\cot \theta = \dfrac{1}{{\tan \theta }}$
$\cos ec\theta = \dfrac{1}{{\sin \theta }}$
$\sec \theta = \dfrac{1}{{\cos \theta }}$
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Complete step-by-step answer:
To prove this equation we will take L.H.S (left hand side) and will convert it into R.H.S (right hand side).
$\dfrac{{{{\tan }^3}x}}{{1 + {{\tan }^2}x}} + \dfrac{{{{\cot }^3}x}}{{1 + {{\cot }^2}x}} = \dfrac{{1 - 2{{\sin }^2}x{{\cos }^2}x}}{{\sin x\cos x}}$
L.H.S= $\dfrac{{{{\tan }^3}x}}{{1 + {{\tan }^2}x}} + \dfrac{{{{\cot }^3}x}}{{1 + {{\cot }^2}x}}$
Now we will apply the identity $1 + {\tan ^2}\theta = {\sec ^2}\theta $ and we will get the following equation.
$ \Rightarrow \dfrac{{{{\tan }^3}x}}{{{{\sec }^2}x}} + \dfrac{{{{\cot }^3}x}}{{1 + {{\cot }^2}x}}$
Now we will apply the identity $1 + {\cot ^2}\theta = \cos e{c^2}\theta $ and we will get the following equation.
$ \Rightarrow \dfrac{{{{\tan }^3}x}}{{{{\sec }^2}x}} + \dfrac{{{{\cot }^3}x}}{{\cos e{c^2}x}}$
Now we will convert $\tan \theta $ into sine and cosine functions by applying relation $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
\[ \Rightarrow \dfrac{{\dfrac{{{{\sin }^3}x}}{{{{\cos }^3}x}}}}{{{{\sec }^2}x}} + \dfrac{{{{\cot }^3}x}}{{\cos e{c^2}x}}\]
Now similarly we will convert $\cot \theta $ into sine and cosine functions by applying relation $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
\[ \Rightarrow \dfrac{{\dfrac{{{{\sin }^3}x}}{{{{\cos }^3}x}}}}{{{{\sec }^2}x}} + \dfrac{{\dfrac{{{{\cos }^3}x}}{{{{\sin }^3}x}}}}{{\cos e{c^2}x}}\]
Now we will apply identity $\sec \theta = \dfrac{1}{{\cos \theta }}$ and $\cos ec\theta = \dfrac{1}{{\sin \theta }}$ to convert whole equation in sine and cosine form because R.H.S is of sine and cosine form.
\[ \Rightarrow \dfrac{{\dfrac{{{{\sin }^3}x}}{{{{\cos }^3}x}}}}{{\dfrac{1}{{{{\cos }^2}x}}}} + \dfrac{{\dfrac{{{{\cos }^3}x}}{{{{\sin }^3}x}}}}{{\dfrac{1}{{{{\sin }^2}x}}}}\]
Now we will write it in a simplified manner so that it is easy to understand.
\[ \Rightarrow \dfrac{{{{\sin }^3}x}}{{{{\cos }^3}x}} \times \dfrac{{{{\cos }^2}x}}{1} + \dfrac{{{{\cos }^3}x}}{{{{\sin }^3}x}} \times \dfrac{{{{\sin }^2}x}}{1}\]
Now we will cancel numerator and denominator equal terms.
\[ \Rightarrow \dfrac{{{{\sin }^3}x}}{{\cos x}} + \dfrac{{{{\cos }^3}x}}{{\sin x}}\]
Now taking L.C.M we will get
\[ \Rightarrow \dfrac{{({{\sin }^3}x)(\sin x) + ({{\cos }^3}x)(\cos x)}}{{\sin x\cos x}}\]
Now we will multiply the following functions in the numerator.
\[ \Rightarrow \dfrac{{({{\sin }^4}x) + ({{\cos }^4}x)}}{{\sin x\cos x}}\]
Now we will apply algebraic identity \[{a^2} + {b^2} = {(a + b)^2} - 2ab\] to further simplify the equation.
\[ \Rightarrow \dfrac{{[{{({{\sin }^2}\theta )}^2} + {{({{\cos }^2}\theta )}^2}]}}{{\sin x\cos x}} = \dfrac{{{{({{\sin }^2}\theta + {{\cos }^2}\theta )}^2} - 2{{\sin }^2}\theta {{\cos }^2}\theta }}{{\sin x\cos x}}\]
Now we will apply identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] and we will get
\[\therefore \dfrac{{1 - 2{{\sin }^2}\theta {{\cos }^2}\theta }}{{\sin x\cos x}}\]
Hence it is proved that $\dfrac{{{{\tan }^3}x}}{{1 + {{\tan }^2}x}} + \dfrac{{{{\cot }^3}x}}{{1 + {{\cot }^2}x}} = \dfrac{{1 - 2{{\sin }^2}x{{\cos }^2}x}}{{\sin x\cos x}}$
Note: Students may likely make mistakes in relating algebraic identities with trigonometric functions but they can be applied easily by only carefully observing the degrees. With the use of algebraic identities questions can be solved with very less complexity. Further this question can be solved by applying identities such as:-
$\tan 3x = \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}$ Now driving another formula from this formula we will get
${\tan ^3}x = 3\tan x - [(\tan 3x)(1 - 3{\tan ^2}x)]$
$\cot 3x = \dfrac{{3\cot x - {{\cot }^3}x}}{{1 - 3{{\cot }^2}x}}$ Similarly we can drive another formula from this equation
${\cot ^3}x = 3\cot x - [(\cot 3x)(1 - 3{\cot ^2}x)]$
Now if we put this two newly derived formula in $\dfrac{{{{\tan }^3}x}}{{1 + {{\tan }^2}x}} + \dfrac{{{{\cot }^3}x}}{{1 + {{\cot }^2}x}}$ then the formed equation will be so complex that it would be very time confusing to solve moreover memorising so many formulas can sometimes became confusing for some students so it is recommended to proceed with the basic method of converting all functions into sine and cosine function.
$1 + {\tan ^2}\theta = {\sec ^2}\theta $
$1 + {\cot ^2}\theta = \cos e{c^2}\theta $
$\cot \theta = \dfrac{1}{{\tan \theta }}$
$\cos ec\theta = \dfrac{1}{{\sin \theta }}$
$\sec \theta = \dfrac{1}{{\cos \theta }}$
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Complete step-by-step answer:
To prove this equation we will take L.H.S (left hand side) and will convert it into R.H.S (right hand side).
$\dfrac{{{{\tan }^3}x}}{{1 + {{\tan }^2}x}} + \dfrac{{{{\cot }^3}x}}{{1 + {{\cot }^2}x}} = \dfrac{{1 - 2{{\sin }^2}x{{\cos }^2}x}}{{\sin x\cos x}}$
L.H.S= $\dfrac{{{{\tan }^3}x}}{{1 + {{\tan }^2}x}} + \dfrac{{{{\cot }^3}x}}{{1 + {{\cot }^2}x}}$
Now we will apply the identity $1 + {\tan ^2}\theta = {\sec ^2}\theta $ and we will get the following equation.
$ \Rightarrow \dfrac{{{{\tan }^3}x}}{{{{\sec }^2}x}} + \dfrac{{{{\cot }^3}x}}{{1 + {{\cot }^2}x}}$
Now we will apply the identity $1 + {\cot ^2}\theta = \cos e{c^2}\theta $ and we will get the following equation.
$ \Rightarrow \dfrac{{{{\tan }^3}x}}{{{{\sec }^2}x}} + \dfrac{{{{\cot }^3}x}}{{\cos e{c^2}x}}$
Now we will convert $\tan \theta $ into sine and cosine functions by applying relation $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
\[ \Rightarrow \dfrac{{\dfrac{{{{\sin }^3}x}}{{{{\cos }^3}x}}}}{{{{\sec }^2}x}} + \dfrac{{{{\cot }^3}x}}{{\cos e{c^2}x}}\]
Now similarly we will convert $\cot \theta $ into sine and cosine functions by applying relation $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
\[ \Rightarrow \dfrac{{\dfrac{{{{\sin }^3}x}}{{{{\cos }^3}x}}}}{{{{\sec }^2}x}} + \dfrac{{\dfrac{{{{\cos }^3}x}}{{{{\sin }^3}x}}}}{{\cos e{c^2}x}}\]
Now we will apply identity $\sec \theta = \dfrac{1}{{\cos \theta }}$ and $\cos ec\theta = \dfrac{1}{{\sin \theta }}$ to convert whole equation in sine and cosine form because R.H.S is of sine and cosine form.
\[ \Rightarrow \dfrac{{\dfrac{{{{\sin }^3}x}}{{{{\cos }^3}x}}}}{{\dfrac{1}{{{{\cos }^2}x}}}} + \dfrac{{\dfrac{{{{\cos }^3}x}}{{{{\sin }^3}x}}}}{{\dfrac{1}{{{{\sin }^2}x}}}}\]
Now we will write it in a simplified manner so that it is easy to understand.
\[ \Rightarrow \dfrac{{{{\sin }^3}x}}{{{{\cos }^3}x}} \times \dfrac{{{{\cos }^2}x}}{1} + \dfrac{{{{\cos }^3}x}}{{{{\sin }^3}x}} \times \dfrac{{{{\sin }^2}x}}{1}\]
Now we will cancel numerator and denominator equal terms.
\[ \Rightarrow \dfrac{{{{\sin }^3}x}}{{\cos x}} + \dfrac{{{{\cos }^3}x}}{{\sin x}}\]
Now taking L.C.M we will get
\[ \Rightarrow \dfrac{{({{\sin }^3}x)(\sin x) + ({{\cos }^3}x)(\cos x)}}{{\sin x\cos x}}\]
Now we will multiply the following functions in the numerator.
\[ \Rightarrow \dfrac{{({{\sin }^4}x) + ({{\cos }^4}x)}}{{\sin x\cos x}}\]
Now we will apply algebraic identity \[{a^2} + {b^2} = {(a + b)^2} - 2ab\] to further simplify the equation.
\[ \Rightarrow \dfrac{{[{{({{\sin }^2}\theta )}^2} + {{({{\cos }^2}\theta )}^2}]}}{{\sin x\cos x}} = \dfrac{{{{({{\sin }^2}\theta + {{\cos }^2}\theta )}^2} - 2{{\sin }^2}\theta {{\cos }^2}\theta }}{{\sin x\cos x}}\]
Now we will apply identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] and we will get
\[\therefore \dfrac{{1 - 2{{\sin }^2}\theta {{\cos }^2}\theta }}{{\sin x\cos x}}\]
Hence it is proved that $\dfrac{{{{\tan }^3}x}}{{1 + {{\tan }^2}x}} + \dfrac{{{{\cot }^3}x}}{{1 + {{\cot }^2}x}} = \dfrac{{1 - 2{{\sin }^2}x{{\cos }^2}x}}{{\sin x\cos x}}$
Note: Students may likely make mistakes in relating algebraic identities with trigonometric functions but they can be applied easily by only carefully observing the degrees. With the use of algebraic identities questions can be solved with very less complexity. Further this question can be solved by applying identities such as:-
$\tan 3x = \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}$ Now driving another formula from this formula we will get
${\tan ^3}x = 3\tan x - [(\tan 3x)(1 - 3{\tan ^2}x)]$
$\cot 3x = \dfrac{{3\cot x - {{\cot }^3}x}}{{1 - 3{{\cot }^2}x}}$ Similarly we can drive another formula from this equation
${\cot ^3}x = 3\cot x - [(\cot 3x)(1 - 3{\cot ^2}x)]$
Now if we put this two newly derived formula in $\dfrac{{{{\tan }^3}x}}{{1 + {{\tan }^2}x}} + \dfrac{{{{\cot }^3}x}}{{1 + {{\cot }^2}x}}$ then the formed equation will be so complex that it would be very time confusing to solve moreover memorising so many formulas can sometimes became confusing for some students so it is recommended to proceed with the basic method of converting all functions into sine and cosine function.
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