Question

Prove the identity $\left( \cos ec\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)=\dfrac{\text{1}}{\text{tan}\theta \text{+cot}\theta }.$

Answer 1

Hint:In the given expression consider the LHS part and RHS part separately. Apply basic trigonometric formulas in $\cos ec\theta ,\sec \theta ,\tan \theta$ and $\cot \theta$ and simplify. Prove that LHS is equal to RHS.

Complete step-by-step answer:
We have been given a trigonometric expression. We need to solve the LHS part of the RHS part. Then prove that LHS is equal to RHS.
We have been given,
$\left( \cos ec\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)=\dfrac{\text{1}}{\text{tan}\theta \text{+cot}\theta }.$
Let us consider the $LHS=\left( \cos ec\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)\to (1)$
We know the basic trigonometric identity,
$\cos ec\theta =\dfrac{\text{1}}{\text{sin}\theta }$ and $\sec \theta =\dfrac{\text{1}}{\text{cos}\theta }.$
Now substitute these value in (1)
$LHS=\left( \cos ec\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)$
$=\left( \dfrac{1}{\sin \theta }-\sin \theta \right)\left( \dfrac{1}{\cos \theta }-\cos \theta \right)$ , let us simplify it
$LHS=\left( \dfrac{1-{{\sin }^{2}}\theta }{\sin \theta } \right)\left( \dfrac{1-{{\cos }^{2}}\theta }{\cos \theta } \right)\to (2)$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
$\therefore {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$ and ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta .$
Thus let us substitute the above expression in (2)
$LHS=\left( \dfrac{{{\cos }^{2}}\theta }{\sin \theta } \right)\times \left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta } \right)=\sin \theta \cos \theta$ .
Now let us consider the $RHS=\dfrac{\text{1}}{\tan \theta +\cot \theta }\to (3)$
We know that by basic trigonometric,
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ .
Now substitute there values in (3) and simplify if,
$RHS=\dfrac{1}{\text{tan}\theta \text{+cot}\theta }$ .
$=\dfrac{\text{1}}{\dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\sin \theta }}$ , simplify the expression.
$=\dfrac{\text{1}}{\dfrac{\text{si}{{\text{n}}^{2}}\theta +{{\cos }^{2}}\theta }{\text{sin}\theta \text{cos}\theta }}=\dfrac{\text{sin}\theta \text{cos}\theta }{\text{si}{{\text{n}}^{2}}\theta +{{\cos }^{2}}\theta }.$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ .
$\Rightarrow RHS=\dfrac{\text{sin}\theta \text{cos}\theta }{\text{1}}=\sin \theta \cos \theta .$
Thus we got $LHS=\sin \theta \cos \theta$ and $RHS=\sin \theta \cos \theta .$
$\therefore LHS=RHS$ .
$\therefore$ We proved that $\left( \cos ec\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)=\dfrac{\text{1}}{\text{tan}\theta \text{+cot}\theta }.$
Thus proved.

Note: You can also solve the RHS part by, $\cot \theta =\dfrac{1}{\tan \theta }$ .
$\dfrac{\text{1}}{\text{tan}\theta \text{+}\dfrac{1}{\tan \theta }}=\dfrac{\text{1}}{\dfrac{\text{ta}{{\text{n}}^{2}}\theta +1}{\text{tan}\theta }}=\dfrac{\text{tan}\theta }{\text{ta}{{\text{n}}^{2}}\theta +1}=\dfrac{\dfrac{\text{sin}\theta }{\text{cos}\theta }}{\dfrac{\text{si}{{\text{n}}^{2}}\theta +{{\cos }^{2}}\theta }{\text{co}{{\text{s}}^{2}}\theta }}$ .
$=\dfrac{\sin \theta }{\cos \theta }\times {{\cos }^{2}}\theta =\sin \theta \cos \theta.$Students should remember the important trigonometric identities and formulas for solving these types of questions.