Question

Prove that \[\left( {1 + \cot \theta - \cos ec\theta } \right)\left( {1 + \tan \theta + \sec \theta } \right) = 2\]

Answer 1

Hint:For all kinds of trigonometric proofs, we need to select any of the sides and reduce all the trigonometric terms in the terms of $\sin \theta $ and $\cos \theta $. Then after simplify the equations and at last we will get the other side form on perfect simplification according to the problem.

Complete step-by-step answer:
In the given problem, the right-hand side of the equation is constant. But we need to take the side where trigonometric terms are contained. In the present problem,
Let us take the left-hand side of the equation and simplify as far as possible.
The left-hand side of the equation is
\[ \Rightarrow \left( {1 + \cot \theta - \cos ec\theta } \right)\left( {1 + \tan \theta + \sec \theta } \right)\]
Convert the $\cos ec\theta $, $\cot \theta $, $\tan \theta $ and $\sec \theta $ in the terms of $\sin \theta $ hand $\cos \theta $.
\[ \Rightarrow \left( {1 + \dfrac{{\cos \theta }}{{\sin \theta }} - \dfrac{1}{{\sin \theta }}} \right)\left( {1 + \dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{1}{{\cos \theta }}} \right)\]
Now, take the \[\sin \theta \] and \[\cos \theta \] as the common denominator for the first and the second terms respectively. So, the equation changes as follows:
\[ \Rightarrow \left( {\dfrac{{(\sin \theta + \cos \theta - 1)}}{{\sin \theta }}} \right)\left( {\dfrac{{(\sin \theta + \cos \theta + 1}}{{\cos \theta }}} \right)\]
When we clearly look at the numerators of the both terms, we can get that it is in the form of $\left( {a + b} \right)\left( {a - b} \right)$
And we know that,
$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$
Comparing the above formula with the equation we obtained we can get that $a = \left( {\sin \theta + \cos \theta } \right)$ and $b = 1$. So, our equation now becomes as follows:
$
   \Rightarrow \dfrac{{{{\left( {\sin \theta + \cos \theta } \right)}^2} - {1^2}}}{{\sin \theta \cos \theta }} \\
    \\
 $
Now in numerator we can see that one part in that is in the form of ${\left( {a + b} \right)^2}$ and we know that
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ where $a = \sin \theta $ and $b = \cos \theta $ as per our problem.
So, applying that we get the equation as,
$ \Rightarrow \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta + 2\sin \theta \cos \theta - 1}}{{\sin \theta \cos \theta }}$
Now clearly, we can point out one of the basic formulae of the trigonometry in the above equations and that is as follows:
${\sin ^2}\theta + {\cos ^2}\theta = 1$
Applying this in the equation we will get,
$
   \Rightarrow \dfrac{{1 + 2\sin \theta \cos \theta - 1}}{{\sin \theta \cos \theta }} \\
   = \dfrac{{2\sin \theta \cos \theta }}{{\sin \theta \cos \theta }} \\
   = 2 \\
 $
After all simplifications, here you can see that we obtained the right-hand side of the equation given in the problem.
As we started from the left-hand side of the equation and reached the right-hand side of the equation, we can say that both the values are equal.
Hence, we proved.

Note:Most probably all the types of trigonometric proofs as we see in the above example can be done in the same method. But there will be some proofs where we need to use the trigonometric formulas strictly. Hence students should remember important trigonometric formulas and identities for solving these types of questions.