Question

Prove: $\left( {1 + \cot \theta - \cos ec\theta } \right)\left( {1 + \tan \theta + \sec \theta } \right) = 2$

Answer 1

Hint:
We can simplify the LHS and prove that it is equal to the RHS. For simplifying the LHS, we can write all the terms in terms of sine and cosines. Then we can take their LCM and simplify. Then we can take the product using the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ . After further simplification using trigonometric identities and cancelling the common terms, we get that LHS is equal to the RHS.

Complete step by step solution:
We need to prove that $\left( {1 + \cot \theta - \cos ec\theta } \right)\left( {1 + \tan \theta + \sec \theta } \right) = 2$ .
We can take the LHS.
 $ \Rightarrow LHS = \left( {1 + \cot \theta - \cos ec\theta } \right)\left( {1 + \tan \theta + \sec \theta } \right)$
We can write all the terms as sin and cos. We know that,
 $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ , $\cos ec\theta = \dfrac{1}{{\sin \theta }}$ , $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\sec \theta = \dfrac{1}{{\cos \theta }}$
On substituting these on the LHS, we get,
 $ \Rightarrow LHS = \left( {1 + \dfrac{{\cos \theta }}{{\sin \theta }} - \dfrac{1}{{\sin \theta }}} \right)\left( {1 + \dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{1}{{\cos \theta }}} \right)$
We can find the LCM of the denominators.
 $ \Rightarrow LHS = \left( {\dfrac{{\sin \theta + \cos \theta - 1}}{{\sin \theta }}} \right)\left( {\dfrac{{\cos \theta + \sin \theta + 1}}{{\cos \theta }}} \right)$
On rearranging, we get,
 $ \Rightarrow LHS = \dfrac{{\left( {\sin \theta + \cos \theta - 1} \right)\left( {\sin \theta + \cos \theta + 1} \right)}}{{\sin \theta \cos \theta }}$
Now the numerator is of the form, $\left( {a + b} \right)\left( {a - b} \right)$ . We know that, $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ .
 $ \Rightarrow LHS = \dfrac{{{{\left( {\sin \theta + \cos \theta } \right)}^2} - {1^2}}}{{\sin \theta \cos \theta }}$
On expanding the squares using the identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ , we get,
 $ \Rightarrow LHS = \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta + 2\sin \theta \cos \theta - 1}}{{\sin \theta \cos \theta }}$
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$ . So, the LHS will become,
 $ \Rightarrow LHS = \dfrac{{1 + 2\sin \theta \cos \theta - 1}}{{\sin \theta \cos \theta }}$
On simplification, we get,
 $ \Rightarrow LHS = \dfrac{{2\sin \theta \cos \theta }}{{\sin \theta \cos \theta }}$
Now we can cancel the common terms.
 $ \Rightarrow LHS = 2$
We have $RHS = 2$ . As the RHS and LHS are equal, we can write,
 $ \Rightarrow LHS = RHS$

Hence proved.

Note:
We must know the following relations to convert all the trigonometric ratios in terms of sines and cosines only.
 $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$
  $\cos ec\theta = \dfrac{1}{{\sin \theta }}$
 $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
 $\sec \theta = \dfrac{1}{{\cos \theta }}$
We must use the algebraic identities for simplification, otherwise the expression will get complex. We cannot start from the RHS and reach the LHS in this problem as the RHS is a constant and cannot be expanded or simplified. We must take care of the signs while applying the identities and while opening the brackets.