Question

Prove the following expression:
\[\cos \theta \left( \tan \theta +2 \right)\left( 2\tan \theta +1 \right)=2\sec \theta +5\sin \theta \]

Answer 1

Hint: To solve this question, we should know the basic concept of trigonometric ratios like \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. Also, we should know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and \[\cos \theta =\dfrac{1}{\sec \theta }\]. By using these relations, we can prove the desired results.

Complete step-by-step answer:

In this question, we are asked to prove that

\[\cos \theta \left( \tan \theta +2 \right)\left( 2\tan \theta +1 \right)=2\sec \theta +5\sin \theta \]

To prove this relation, we will first consider the left-hand side of the relation. So, we can write it as

\[LHS=\cos \theta \left( \tan \theta +2 \right)\left( 2\tan \theta +1 \right)\]

Now, we know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. So by using this relation, we can write LHS as

\[LHS=\cos \theta \left( \dfrac{\sin \theta }{\cos \theta }+2 \right)\left( \dfrac{2\sin \theta }{\cos \theta }+1 \right)\]

Now we will take the LCM of the terms of the LHS, so we will get

\[LHS=\cos \theta \left[ \dfrac{\sin \theta +2\cos \theta }{\cos \theta } \right]\left[ \dfrac{2\sin \theta +\cos \theta }{\cos \theta } \right]\]

Now we will simplify it further, so we will get

\[LHS=\cos \theta \left[ \dfrac{2{{\sin }^{2}}\theta +4\sin \theta \cos \theta +\sin \theta \cos \theta +2{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta } \right]\]

Now we will add similar terms, so we get

\[LHS=\cos \theta \left[ \dfrac{2\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+5\sin \theta \cos \theta }{{{\cos }^{2}}\theta } \right]\]

Now we can see that \[\cos \theta \] is common in both the numerator and denominator, so we will cancel them with each other and we will get

\[LHS=\dfrac{2\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+5\sin \theta \cos \theta }{\cos \theta }\]

Now we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. So, we will get

\[LHS=\dfrac{2+5\sin \theta \cos \theta }{\cos \theta }\]

Now we will separate the terms of the numerator. So, we will get,

\[LHS=\dfrac{2}{\cos \theta }+\dfrac{5\sin \theta \cos \theta }{\cos \theta }\]

Now, we know that \[\cos \theta =\dfrac{1}{\sec \theta }\]. So, we can say \[\sec \theta =\dfrac{1}{\cos \theta }\]. Also, we can see that \[\cos \theta \] is common in numerator and denominator of the second term of the LHS. So, we will get

\[LHS=2\sec \theta +5\sin \theta \]

LHS = RHS

Hence proved


Note: In this question, we can directly open brackets, and then we can apply the formulas like \[{{\tan }^{2}}\theta +1={{\sec }^{2}}\theta \], \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and \[\cos \theta =\dfrac{1}{\sec \theta }\]. By using these formulas, we can also derive the desired results.