Question

$\tan {{63}^{{}^\circ }}+\cos {{22}^{{}^\circ }}=\cot \theta +\sin {{68}^{{}^\circ }}$ , then $\theta =$
(a) ${{30}^{{}^\circ }}$
(b) ${{27}^{{}^\circ }}$
(c) \[{{37}^{{}^\circ }}\]
(d) ${{27}^{{}^\circ }}$

Answer 1

Hint: To solve this question, we will use the concept of values of trigonometric function, which are $\tan ({{90}^{{}^\circ }}-\theta )=\cot \theta $ and $cos({{90}^{{}^\circ }}-\theta )=\sin \theta $. After evaluating the values of left hand side functions, we will substitute the values in equation $\tan {{63}^{{}^\circ }}+\cos {{22}^{{}^\circ }}=\cot \theta +\sin {{68}^{{}^\circ }}$and by comparing, we will get values of $\theta $ .

Complete step-by-step answer:
Before we solve the question, let's see some important angle formulas for trigonometric values. This formula tells how we can write trigonometric functions in terms of another trigonometric function. So, these formulas are,
$\tan ({{90}^{{}^\circ }}-\theta )=\cot \theta $
$cos({{90}^{{}^\circ }}-\theta )=\sin \theta $
And, $cosec({{90}^{{}^\circ }}-\theta )=\sec \theta $

In question we are given that $\tan {{63}^{{}^\circ }}+\cos {{22}^{{}^\circ }}=\cot \theta +\sin {{68}^{{}^\circ }}$….( i )
As we discussed above that $\tan ({{90}^{{}^\circ }}-\theta )=\cot \theta $ .
So, we can write $\tan {{63}^{{}^\circ }}$ as $\tan ({{90}^{{}^\circ }}-{{27}^{{}^\circ }})$, as 90 – 27 = 63 .
Also, we discussed that $cos({{90}^{{}^\circ }}-\theta )=\sin \theta $ ,
So, also we can write $\cos {{22}^{{}^\circ }}$ as $cos({{90}^{{}^\circ }}-{{68}^{{}^\circ }})$, as 90 – 68 = 22 .
Substituting, values of $\tan {{63}^{{}^\circ }}$ as $\tan ({{90}^{{}^\circ }}-{{27}^{{}^\circ }})$ and in equation ( i ), we get
$\tan ({{90}^{{}^\circ }}-{{27}^{{}^\circ }})+\cos {{22}^{{}^\circ }}=\cot \theta +\sin {{68}^{{}^\circ }}$…..( ii )
Substituting, values of $\cos {{22}^{{}^\circ }}$ as $cos({{90}^{{}^\circ }}-{{68}^{{}^\circ }})$ in equation ( ii ), we get
$\tan ({{90}^{{}^\circ }}-{{27}^{{}^\circ }})+cos({{90}^{{}^\circ }}-{{68}^{{}^\circ }})=\cot \theta +\sin {{68}^{{}^\circ }}$…..( iii )
Also, we discussed above that $cos({{90}^{{}^\circ }}-\theta )$ is equals to $sin\theta $ that is $cos({{90}^{{}^\circ }}-\theta )=sin\theta $
So, we can write $cos({{90}^{{}^\circ }}-{{68}^{{}^\circ }})$ as $\sin {{68}^{{}^\circ }}$, that is $cos({{90}^{{}^\circ }}-{{68}^{{}^\circ }})=\sin {{68}^{{}^\circ }}$
And Also, we discussed above that $tan({{90}^{{}^\circ }}-\theta )$ is equals to $\cot \theta $ that is $tan({{90}^{{}^\circ }}-\theta )=\cot \theta $
So, we can write $\tan ({{90}^{{}^\circ }}-{{27}^{{}^\circ }})$ as $\cot {{27}^{{}^\circ }}$, that is $\tan ({{90}^{{}^\circ }}-{{27}^{{}^\circ }})=\cot {{27}^{{}^\circ }}$
Substituting, $\tan ({{90}^{{}^\circ }}-{{27}^{{}^\circ }})$ as $\cot {{27}^{{}^\circ }}$ in ( iii ), we get
$\cot {{27}^{{}^\circ }}+cos({{90}^{{}^\circ }}-{{68}^{{}^\circ }})=\cot \theta +\sin {{68}^{{}^\circ }}$….( iv )
And, substituting $cos({{90}^{{}^\circ }}-{{68}^{{}^\circ }})$ as $\sin {{68}^{{}^\circ }}$in equation ( v ), we get
$\cot {{27}^{{}^\circ }}+\sin {{68}^{{}^\circ }}=\cot \theta +\sin {{68}^{{}^\circ }}$
Now, as on both Left hand side and Right hand side we have same functions, so we can compare the inputs of the functions,
So, on comparison, we get
$\cot {{27}^{{}^\circ }}=\cot \theta $
So, $\theta ={{27}^{{}^\circ }}$

So, the correct answer is “Option b”.

Note: To solve these type of questions, one must know the different angle formulas $\tan ({{90}^{{}^\circ }}-\theta )=\cot \theta $, $cos({{90}^{{}^\circ }}-\theta )=\sin \theta $ and $cosec({{90}^{{}^\circ }}-\theta )=\sec \theta $. While using these formulas evaluate the value of ${{90}^{{}^\circ }}-\theta $carefully as this may change the question. Try to avoid calculation mistakes.