Question

(1 − tan² 45°) / (1 + tan² 45°)
(a) tan 90° (b) 1 (c) sin 45° (d) 0

Answer 1

Answer:

Option D is correct.

Step-by-step explanation:

To solve

$\frac{1 - {tan}^{2}45^ \circ }{1 + {tan}^{2}45^ \circ} \\ \\ \because \: tan \: 45^ \circ = 1$

We know that

${(tan45^ \circ \:})^{2} = {tan}^{2} 45^ \circ \\ \\ so \: {(tan45^ \circ \:})^{2} = {tan}^{2} 45^ \circ = 1$

put the value in the given expression

$\frac{1 - {tan}^{2}45^ \circ }{1 + {tan}^{2}45^ \circ} = \frac{1 - 1}{1 + 1} \\ \\ \implies \frac{0}{2} \\ \\ \frac{1 - {tan}^{2}45^ \circ }{1 + {tan}^{2}45^ \circ} = 0$

because zero by something is equal to zero .

So,option D is correct.

Hope it helps you.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{ \frac{1 - { \tan }^{2} \theta }{1 + { \tan }^{2} \theta} \: = \cos2 \theta\: }$

CALCULATION

USING FORMULA

$\displaystyle \sf{ \frac{1 - { \tan }^{2} {45}^{ \circ} }{1 + { \tan }^{2} {45}^{ \circ}} \: \: }$

$= \sf{ \cos(2 \times {45}^{ \circ} ) \: }$

$= \sf{ \cos {90}^{ \circ} \: }$

$= \sf{ 0\: }$

USING VALUES

$\displaystyle \sf{ \frac{1 - { \tan }^{2} {45}^{ \circ} }{1 + { \tan }^{2} {45}^{ \circ}} \: \: }$

$= \displaystyle \sf{ \frac{1 - {(1)}^{2} }{1 + {(1)}^{2} } \: \: }$

$= \displaystyle \sf{ \frac{0}{2} \: }$

$= \displaystyle \sf{ 0 \: }$

RESULT

$\boxed{ \: \: \displaystyle \sf{ \frac{1 - { \tan }^{2} {45}^{ \circ} }{1 + { \tan }^{2} {45}^{ \circ}} \: = 0 \: \: \: }}$

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