Question

Evaluate the following expression and find out its value
\[\cot {{36}^{o}}-\tan {{54}^{o}}\].

Answer 1

Hint: This question is based on trigonometric relations. Try to recall the relations between \[\cot \theta \] and \[\tan \theta \] involving complementary angles.

Complete step-by-step answer:

Let us consider two angles \[\alpha \] and \[\beta \] such that \[\alpha +\beta ={{90}^{o}}\]. Here , \[\alpha \] and \[\beta \] are called complementary angles .
Now, we know , \[\sin \left( {{90}^{o}}-\theta \right)=\cos \theta \] and \[\cos \left( {{90}^{o}}-\theta \right)=\sin \theta \].
Now , when we consider the angles \[\alpha \] and \[\beta \] such that \[\alpha +\beta ={{90}^{o}}\] , we get \[\alpha ={{90}^{o}}-\beta \].
So , \[\sin \alpha =\sin \left( {{90}^{o}}-\beta \right)=\cos \beta ......(i)\] and \[\cos \alpha =\cos \left( {{90}^{o}}-\beta \right)=\sin \beta .....(ii)\].
Now , we know , \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\].
So , \[\cot \alpha =\dfrac{\cos \alpha }{\sin \alpha }....(iii)\].
Now , we will substitute equations \[(i)\] and \[(ii)\]in equation\[(iii)\].
On substituting equations \[(i)\] and \[(ii)\]in equation\[(iii)\], we get
\[\cot \alpha =\dfrac{\cos \left( {{90}^{o}}-\beta \right)}{\sin \left( {{90}^{o}}-\beta \right)}\]
So , \[\cot \alpha =\dfrac{\sin \beta }{\cos \beta }\]
Or , \[\cot \alpha =\dfrac{1}{\cot \beta }.....(iv)\]
Now , we know \[\cot \beta =\dfrac{1}{\tan \beta }\].
We will take \[\tan \beta \] to the left hand side of the equation and \[\cot \beta \] to the right hand side of the equation .
So , we get \[\tan \beta =\dfrac{1}{\cot \beta }.....(v)\].
Now , we will substitute equation \[(v)\] in equation\[(iv)\].
On substituting equation \[(v)\] in equation\[(iv)\], we get \[\cot \alpha =\tan \beta ...(vi)\].
Now , we will consider the value of \[\alpha \] to be equal to \[{{36}^{o}}\].
Now , we know , \[\beta ={{90}^{o}}-\alpha \] .
So , when the value of \[\alpha \] is equal to \[{{36}^{o}}\], then the value of \[\beta \] is given as \[\beta ={{90}^{o}}-{{36}^{o}}={{54}^{o}}\].
Now , from equation\[(vi)\], we have \[\cot \alpha =\tan \beta \].
We will substitute the values of \[\alpha \] and \[\beta \] in equation\[(vi)\].
On substituting the values of \[\alpha \] and \[\beta \] in equation\[(vi)\], we get \[\cot {{36}^{o}}=\tan {{54}^{o}}....(vii)\].
Now , in the question we are asked to find the value of \[\cot {{36}^{o}}-\tan {{54}^{o}}\].
But , from equation\[(vii)\], we know that \[\cot {{36}^{o}}=\tan {{54}^{o}}\].
So , \[\cot {{36}^{o}}-\tan {{54}^{o}}=\cot {{36}^{o}}-\cot {{36}^{o}}=0\].
Hence , the value of \[\cot {{36}^{o}}-\tan {{54}^{o}}\] is equal to \[0\].

Note: Students generally get confused between \[\cos \left( {{90}^{o}}-\theta \right)=\sin \theta \] and \[\cos \left( {{90}^{o}}+\theta \right)=-\sin \theta \].
Sometimes , by mistake students write \[\cos \left( {{90}^{o}}-\theta \right)\] as \[-\sin \theta \] and \[\cos \left( {{90}^{o}}+\theta \right)\] as \[\sin \theta \], which is wrong . Sign mistakes are common but can result in getting a wrong answer . Hence , students should be careful while using trigonometric formulae and should take care of sign conventions.