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Question: The value of \(\dfrac{1-{{\tan }^{2}}{{45}^{\circ }}}{1+{{\tan }^{2}}{{45}^{\circ }}}\) is equal to:...

The value of 1−tan⁡245∘1+tan⁡245∘\dfrac{1-{{\tan }^{2}}{{45}^{\circ }}}{1+{{\tan }^{2}}{{45}^{\circ }}} is equal to:
(a). tan⁡90∘\tan {{90}^{\circ }}
(b). sin⁡45∘\sin {{45}^{\circ }}
(c). 1
(d). sin⁡0∘\sin {{0}^{\circ }}

Explanation

Solution

In the given expression, if you know the value of tan⁡450\tan {{45}^{0}} then you are done with the question. From the trigonometric functions, we know the value of tan⁡450\tan {{45}^{0}} is equal to 1 so substitute the value of tan⁡450\tan {{45}^{0}} and solve.

Complete step-by-step answer:
The expression given in the question is:
1−tan⁡245∘1+tan⁡245∘\dfrac{1-{{\tan }^{2}}{{45}^{\circ }}}{1+{{\tan }^{2}}{{45}^{\circ }}}
The thing that makes the above problem very easy is the value of tan⁡450\tan {{45}^{0}} . From the trigonometric functions, the value of tan⁡450=1\tan {{45}^{0}}=1 so substitute this value of tan⁡450\tan {{45}^{0}} in the above expression and we get,
1−(1)21+(1)2 =0 \begin{aligned} & \dfrac{1-{{\left( 1 \right)}^{2}}}{1+{{\left( 1 \right)}^{2}}} \\\ & =0 \\\ \end{aligned}
From the above calculation, we have found the given expression as 0.
Now, we are going to find the value of each option and compare with the result of each option with the result of the given expression.
(a) tan⁡90∘=∞\tan {{90}^{\circ }}=\infty
The answer of the above option is not equal to the result of the given expression.
(b) sin⁡45∘=12\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}
The answer of the above option is not equal to the result of the given expression.
(c) 1
The above option is not equal to the answer of the given expression.
(d) sin⁡0∘=0\sin {{0}^{\circ }}=0
The answer of the above option is equal to the answer of the given expression.
From the above discussion of the options given in the question, we have found that the answer of option (d) is equal to the given expression.
Hence, the correct option is (d).

Note: The other way of solving the above question is as follows:
The expression given in the question is:
1−tan⁡245∘1+tan⁡245∘\dfrac{1-{{\tan }^{2}}{{45}^{\circ }}}{1+{{\tan }^{2}}{{45}^{\circ }}}
The above expression is the expansion of the trigonometric identity:
cos⁡(2(45∘))=1−tan⁡245∘1+tan⁡245∘\cos \left( 2\left( {{45}^{\circ }} \right) \right)=\dfrac{1-{{\tan }^{2}}{{45}^{\circ }}}{1+{{\tan }^{2}}{{45}^{\circ }}}
From the above equation we can write the given expression as:
1−tan⁡245∘1+tan⁡245∘=cos⁡90∘\dfrac{1-{{\tan }^{2}}{{45}^{\circ }}}{1+{{\tan }^{2}}{{45}^{\circ }}}=\cos {{90}^{\circ }}
From the values of trigonometric functions we know that,
cos⁡90∘=0\cos {{90}^{\circ }}=0
Hence, the value of the given expression is 0 and the option (d) has the value equals 0.