Answer
Verified
425.1k+ views
Hint: Divide numerator and denominator by $\cos {{9}^{\circ }}$ or $\sin {{9}^{\circ }}$ to get a trigonometric identity.
Complete step-by-step answer:
Use relation $\cot \left( 90-\theta \right)=\tan \theta $ or $\tan \left( 90-\theta \right)=\cot \theta $ for simplifying it further.
We have to prove that,
$\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}=\cot {{36}^{\circ }}..........\left( i \right)$
Let us prove the given relation by simplifying LHS of equation (i) we have,
LHS$=\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}$
Dividing numerator and denominator by $\cos {{9}^{\circ }}$, we get,
LHS = $\dfrac{\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}{\dfrac{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}$
Now, we can divide $\cos {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ and $\sin {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ in numerator and similarly, $\cos {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ and $\sin {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ in denominator as well. We get,
LHS = $\dfrac{\dfrac{\cos {{9}^{\circ }}}{\cos {{9}^{\circ }}}+\dfrac{\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}{\dfrac{\cos {{9}^{\circ }}}{\cos {{9}^{\circ }}}-\dfrac{\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}............(ii)$
We have trigonometric identity,
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }...........(iii)$
Hence, equation (ii) can be simplified as
LHS = $\dfrac{1+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}}=\dfrac{1+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}\left( 1 \right)}$
As we know, $\tan {{45}^{\circ }}=1$ , so above equation can be written as
LHS = $\dfrac{\tan {{45}^{\circ }}+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}\tan {{45}^{\circ }}}.............\left( iv \right)$
Now, using trigonometric identity,
$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A+\tan B}$
Hence equation (iv) can be rewritten with the help of the above equation. So, we get
LHS = $\tan \left( 45+9 \right)=\tan {{54}^{\circ }}$
Now, we can use relation $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $ to convert the ‘tan’ function to ‘cot’.
Hence, LHS can be given as
LHS = $\tan \left( 90-36 \right)=\cot {{36}^{\circ }}$
Therefore, LHS = RHS = $\cot {{36}^{\circ }}$
Hence proved the given relation
Note: Another approach for the given question would be to use trigonometric identities
$\cos C+\cos D=2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$
$\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$
Use relation $\cos \theta =\sin \left( 90-\theta \right)$
So, LHS = $\dfrac{\cos {{9}^{\circ }}+\cos {{81}^{\circ }}}{\cos 9-\cos 81}$
One can go wrong while converting $\dfrac{\tan 45+\tan {{9}^{\circ }}}{1-\tan 9\tan 45}$ to $\tan {{54}^{\circ }}$ .
He or she may confuse between formula tan(A-B) and tan(A+B).
Complete step-by-step answer:
Use relation $\cot \left( 90-\theta \right)=\tan \theta $ or $\tan \left( 90-\theta \right)=\cot \theta $ for simplifying it further.
We have to prove that,
$\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}=\cot {{36}^{\circ }}..........\left( i \right)$
Let us prove the given relation by simplifying LHS of equation (i) we have,
LHS$=\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}$
Dividing numerator and denominator by $\cos {{9}^{\circ }}$, we get,
LHS = $\dfrac{\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}{\dfrac{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}$
Now, we can divide $\cos {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ and $\sin {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ in numerator and similarly, $\cos {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ and $\sin {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ in denominator as well. We get,
LHS = $\dfrac{\dfrac{\cos {{9}^{\circ }}}{\cos {{9}^{\circ }}}+\dfrac{\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}{\dfrac{\cos {{9}^{\circ }}}{\cos {{9}^{\circ }}}-\dfrac{\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}............(ii)$
We have trigonometric identity,
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }...........(iii)$
Hence, equation (ii) can be simplified as
LHS = $\dfrac{1+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}}=\dfrac{1+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}\left( 1 \right)}$
As we know, $\tan {{45}^{\circ }}=1$ , so above equation can be written as
LHS = $\dfrac{\tan {{45}^{\circ }}+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}\tan {{45}^{\circ }}}.............\left( iv \right)$
Now, using trigonometric identity,
$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A+\tan B}$
Hence equation (iv) can be rewritten with the help of the above equation. So, we get
LHS = $\tan \left( 45+9 \right)=\tan {{54}^{\circ }}$
Now, we can use relation $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $ to convert the ‘tan’ function to ‘cot’.
Hence, LHS can be given as
LHS = $\tan \left( 90-36 \right)=\cot {{36}^{\circ }}$
Therefore, LHS = RHS = $\cot {{36}^{\circ }}$
Hence proved the given relation
Note: Another approach for the given question would be to use trigonometric identities
$\cos C+\cos D=2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$
$\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$
Use relation $\cos \theta =\sin \left( 90-\theta \right)$
So, LHS = $\dfrac{\cos {{9}^{\circ }}+\cos {{81}^{\circ }}}{\cos 9-\cos 81}$
One can go wrong while converting $\dfrac{\tan 45+\tan {{9}^{\circ }}}{1-\tan 9\tan 45}$ to $\tan {{54}^{\circ }}$ .
He or she may confuse between formula tan(A-B) and tan(A+B).
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE