Question

Prove the following:
\[\dfrac{\cos \left( {{90}^{o}}-\theta \right)\sec \left( {{90}^{o}}-\theta \right)\tan \theta }{\operatorname{cosec}\left( {{90}^{o}}-\theta \right)\sin \left( {{90}^{o}}-\theta \right)\cot \left( {{90}^{o}}-\theta \right)}+\dfrac{\tan \left( {{90}^{o}}-\theta \right)}{\cot \theta }=2\]

Answer 1

Hint: First of all, take the LHS of the given question and use sin A.cosec A = 1 and cos A.sec A = 1 to simplify the given expression. To further simplify it, use \[\cot \left( {{90}^{o}}-A \right)=\tan A\] and \[\tan \left( {{90}^{o}}-A \right)=\cot A\]. Finally, cancel the like terms to get the desired answer.

Complete step-by-step answer:
Here, we have to prove that
\[\dfrac{\cos \left( {{90}^{o}}-\theta \right)\sec \left( {{90}^{o}}-\theta \right)\tan \theta }{\operatorname{cosec}\left( {{90}^{o}}-\theta \right)\sin \left( {{90}^{o}}-\theta \right)\cot \left( {{90}^{o}}-\theta \right)}+\dfrac{\tan \left( {{90}^{o}}-\theta \right)}{\cot \theta }=2\]
Let us consider the left-hand side (LHS) of the equation given in the question as
\[L=\dfrac{\cos \left( {{90}^{o}}-\theta \right)\sec \left( {{90}^{o}}-\theta \right)\tan \theta }{\operatorname{cosec}\left( {{90}^{o}}-\theta \right)\sin \left( {{90}^{o}}-\theta \right)\cot \left( {{90}^{o}}-\theta \right)}+\dfrac{\tan \left( {{90}^{o}}-\theta \right)}{\cot \theta }\]
We know that cos A.sec A = 1. By substituting \[A={{90}^{o}}-\theta \], we get, \[\cos \left( {{90}^{o}}-\theta \right).\sec \left( {{90}^{o}}-\theta \right)=1\]. Using this in the above expression, we get,
\[L=\dfrac{1.\tan \theta }{\operatorname{cosec}\left( {{90}^{o}}-\theta \right)\sin \left( {{90}^{o}}-\theta \right)\cot \left( {{90}^{o}}-\theta \right)}+\dfrac{\tan \left( {{90}^{o}}-\theta \right)}{\cot \theta }\]
Now, we also know that cosec A.sin A = 1. By substituting \[A={{90}^{o}}-\theta \], we get, \[\operatorname{cosec}\left( {{90}^{o}}-\theta \right).\sin \left( {{90}^{o}}-\theta \right)=1\]. Using this in the above expression, we get,
\[L=\dfrac{1.\tan \theta }{1.\cot \left( {{90}^{o}}-\theta \right)}+\dfrac{\tan \left( {{90}^{o}}-\theta \right)}{\cot \theta }\]
We know that \[\cot \left( {{90}^{o}}-\theta \right)=\tan A\]. By using this in the above expression, we get,
\[L=\dfrac{\tan \theta }{\tan \theta }+\dfrac{\tan \left( {{90}^{o}}-\theta \right)}{\cot \theta }\]
Also, we know that \[\tan \left( {{90}^{o}}-\theta \right)=\cot A\]. By using this in the above expression, we get,
\[L=\dfrac{\tan \theta }{\tan \theta }+\dfrac{\cot \theta }{\cot \theta }\]
By canceling the like terms in the above expression, we get L = 1 + 1 = 2 which is equal to RHS of the equation given in the question.
So, we get LHS. RHS. Hence proved.
Therefore, we have proved that, \[\dfrac{\cos \left( {{90}^{o}}-\theta \right)\sec \left( {{90}^{o}}-\theta \right)\tan \theta }{\operatorname{cosec}\left( {{90}^{o}}-\theta \right)\sin \left( {{90}^{o}}-\theta \right)\cot \left( {{90}^{o}}-\theta \right)}+\dfrac{\tan \left( {{90}^{o}}-\theta \right)}{\cot \theta }=2\]

Note: Students should remember the general formulas like
 \[\begin{align}
  & \sin \left( {{90}^{o}}-\theta \right)=\cos \theta ,\text{ cos}\left( {{90}^{o}}-\theta \right)=\sin \theta ,\text{ } \\
 & \sec \left( {{90}^{o}}-\theta \right)=\operatorname{cosec}\theta ,\text{ cosec}\left( {{90}^{o}}-\theta \right)=\sec \theta , \\
 & \tan \left( {{90}^{o}}-\theta \right)=\cot \theta \text{ and cot}\left( {{90}^{o}}-\theta \right)=\tan \theta \\
\end{align}\]
Also, some students often get confused if \[\sin \left( {{90}^{o}}-\theta \right).\text{cosec}\left( {{90}^{o}}-\theta \right)\] and \[\sec \left( {{90}^{o}}-\theta \right).\text{cos}\left( {{90}^{o}}-\theta \right)\] are equal to 1 or not. So they must note that sin A.cosec A and sec B.cos B are also equal to 1 whatever is the value of angle A and B.