Question

Choose the correct option and justify your choice;
$\dfrac{{1 - {{\tan }^2}{{45}^0}}}{{1 + {{\tan }^2}{{45}^0}}} =$
$a.{\text{ }}\tan {90^0} \\\ b.{\text{ 1}} \\\ c.{\text{ }}\sin {45^0} \\\ d.{\text{ 0}} \\\$

Answer 1

Hint – In this question use trigonometric identities which is $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and ${\sin ^2}\theta + {\cos ^2}\theta = 1{\text{, }}{\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta$, to reach the answer.

Given equation is
$\dfrac{{1 - {{\tan }^2}{{45}^0}}}{{1 + {{\tan }^2}{{45}^0}}}$
Method - 1
As we know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Substitute this value in above equation we have
$\dfrac{{1 - \dfrac{{{{\sin }^2}{{45}^0}}}{{{{\cos }^2}{{45}^0}}}}}{{1 + \dfrac{{{{\sin }^2}{{45}^0}}}{{{{\cos }^2}{{45}^0}}}}} = \dfrac{{{{\cos }^2}{{45}^0} - {{\sin }^2}{{45}^0}}}{{{{\cos }^2}{{45}^0} + {{\sin }^2}{{45}^0}}}$
Now as we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1{\text{, }}{\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta$
$\Rightarrow \dfrac{{{{\cos }^2}{{45}^0} - {{\sin }^2}{{45}^0}}}{{{{\cos }^2}{{45}^0} + {{\sin }^2}{{45}^0}}} = \dfrac{{\cos \left( {2 \times {{45}^0}} \right)}}{1} = \cos {90^0}$
Now we all know that the value of $\cos {90^0}$ is zero.
$\dfrac{{1 - {{\tan }^2}{{45}^0}}}{{1 + {{\tan }^2}{{45}^0}}} = \cos {90^0} = 0$

Method – 2
As we all know that the value of $\tan {45^0}$ is one.
$\Rightarrow \dfrac{{1 - {{\tan }^2}{{45}^0}}}{{1 + {{\tan }^2}{{45}^0}}} = \dfrac{{1 - 1}}{{1 + 1}} = \dfrac{0}{2} = 0$
Hence, option (d) is correct.

Note – In such types of questions the key concept we have to remember is that always recall the basic trigonometric identities which are stated above and always remember the values of all standard angles, so apply these properties and values in the given equation we will get the required answer.