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# Prove the trigonometric identity: $\dfrac{\cos {{11}^{\circ }}-\sin {{11}^{\circ }}}{\cos {{11}^{\circ }}+\sin {{11}^{\circ }}}=\cot {{56}^{\circ }}$

Last updated date: 16th Jul 2024
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Hint: Convert cosine functions involved in the expression to sine functions using the identity $\cos \left( 90-\theta \right)=\sin \theta .$ Now, use identities of $\sin C-\sin D$ and $\sin C+\sin D$ to get the Right hand side.

Complete step by step solution:
We have to prove,
$\dfrac{\cos {{11}^{\circ }}-\sin {{11}^{\circ }}}{\cos {{11}^{\circ }}+\sin {{11}^{\circ }}}=\cot 56.................\left( i \right)$
As we need to prove the above equation, we can simplify the left hand side of the equation to equate it to the right hand side.
$LHS=\dfrac{\cos {{11}^{\circ }}-\sin {{11}^{\circ }}}{\cos {{11}^{\circ }}+\sin {{11}^{\circ }}}.................\left( ii \right)$
Now we can make the whole equation is sine function only by converting cosine to sine by using the formula,
$\cos \left( 90-\theta \right)=\sin \theta ..................\left( iii \right)$
So, we can write $\cos {{11}^{\circ }}$ as $\cos \left( {{90}^{\circ }}-{{79}^{\circ }} \right)$ and hence, using the equation (iii) we get,
$\cos {{11}^{\circ }}=\cos \left( {{90}^{\circ }}-{{79}^{\circ }} \right)=\sin {{79}^{\circ }}..............\left( iv \right)$
Now replace $\cos {{11}^{\circ }}$ from the equation (ii) by using equation (iv). Hence we get,
$LHS=\dfrac{\sin {{79}^{\circ }}-\sin {{11}^{\circ }}}{\sin {{79}^{\circ }}+\sin {{11}^{\circ }}}.................\left( v \right)$
Now, we can use trigonometric identity of $\sin A-\sin B$ and $\sin A+\sin B$ to evaluate the value of expression in equation (v). Hence, identities of $\sin A-\sin B$ and $\sin A+\sin B$ can be given as,
$\sin A-\sin B=2\sin \dfrac{A-B}{2}\cos \dfrac{A+B}{2}$
$\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$
Hence, we can simplify equation (V) by using the above identities. Hence, we get
$LHS=\dfrac{2\sin \left( \dfrac{{{79}^{\circ }}-{{11}^{\circ }}}{2} \right)\cos \left( \dfrac{{{79}^{\circ }}+{{11}^{\circ }}}{2} \right)}{2\sin \left( \dfrac{{{79}^{\circ }}+{{11}^{\circ }}}{2} \right)\cos \left( \dfrac{{{79}^{\circ }}-{{11}^{\circ }}}{2} \right)}$
$LHS=\dfrac{\sin \left( \dfrac{68}{2} \right)\cos \left( \dfrac{90}{2} \right)}{\sin \left( \dfrac{90}{2} \right)\cos \left( \dfrac{68}{2} \right)}$
$LHS=\dfrac{\sin {{34}^{\circ }}\cos {{45}^{\circ }}}{\sin {{45}^{\circ }}\cos {{34}^{\circ }}}$
Now, we can put values of $\sin {{45}^{\circ }}$ and $\cos {{45}^{\circ }}$ as $\dfrac{1}{\sqrt{2}}.$ Hence we get
$LHS=\dfrac{\sin {{34}^{\circ }}\left( \dfrac{1}{\sqrt{2}} \right)}{\left( \dfrac{1}{\sqrt{2}} \right)\cos {{34}^{\circ }}}=\dfrac{\sin {{34}^{\circ }}}{\cos {{34}^{\circ }}}$
Now, we know that $\dfrac{\sin \theta }{\cos \theta }=\tan \theta .$ Hence we get LHS as
$LHS=\tan {{34}^{\circ }}................\left( vi \right)$
Now, we can convert the $'\tan '$expression of LHS to cosine by using the identity
$\tan \left( 90-\theta \right)=\cot \theta .............\left( vii \right)$
Hence, we can write the LHS from equation (vii), we get
$LHS=\tan {{34}^{\circ }}=\tan \left( 90-56 \right)$
$LHS=\cot {{56}^{\circ }}..............\left( viii \right)$
Hence, from the equation (i) and (viii) we get,
$LHS=RHS=\cot {{56}^{\circ }}$
So, the given expression is proved.

Note: One can convert $\sin {{11}^{\circ }}$ to cosine function as well by using the relation $\sin \left( 90-\theta \right)=\cos \theta .$ And hence apply formula of $\cos A-\cos B$ and $\cos A+\cos B$ to get the answer.
We can divide the whole expression by $\cos {{11}^{\circ }}$ i.e. numerator and denominator both. Hence we get
$\dfrac{1-\tan {{11}^{\circ }}}{1+\tan {{11}^{\circ }}}=\dfrac{\tan {{45}^{\circ }}-\tan {{11}^{\circ }}}{1+\tan {{45}^{\circ }}\tan {{11}^{\circ }}}=\tan \left( 45-11 \right)=\tan {{34}^{\circ }}=\cot {{56}^{\circ }}=RHS.$
Always remember the identities of trigonometric we need to use them according to the question for the flexibility of solution. Don’t get confused with the formula of $\sin A-\sin B$ and $\sin A+\sin B$ in the solution.