Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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}} \\$

Last updated date: 20th Jul 2024
Total views: 452.4k
Views today: 6.52k
Verified
452.4k+ views
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.