Question

8. If 3 cot A = 4, check whether \frac{1-tan^{2}A}{1+tan^{2}A}
1+tan
2
A
1−tan
2
A
​
= cos2 A – sin 2 A or not.

Answer 1

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⭐${\black{\huge{Explanation}}}$⭐

Let ∆ ABC in which B=90°

Let ∆ ABC in which B=90°We know that , cot function is the reciprocal of tan function and it is written as

$cot(a) = \frac{ab}{bc} = \frac{4}{3}$

Let AB = 4k and BC = 3k, where k is a positive real number .

Let AB = 4k and BC = 3k, where k is a positive real number .According to the pythagoream theorem.

${ac}^{2} = {ab }^{2} + {bc}^{2} \\ {ac}^{2} = (4k) {}^{2} + (3k) {}^{2} \\ ac {}^{2} = 16k {}^{2} + 9k {}^{2} \\ ac {}^{2} = 25k {}^{2} \\ ac = 5k$

Now apply the values corresponding to the ratios .

$\tan(a) = \frac{bc}{ab } = \frac{3}{4} \\ \sin(a) = \frac{bc}{ac} = \frac{3}{5} \\ \cos(a) = \frac{ab}{ac} = \frac{4}{5}$

Now compare the left hand side (LHS) with the right hand side (RHS)

LHS

$= \frac{1 - tan {}^{2} a }{1 + tan {}^{2}a } = \frac{1 - ( \frac{2}{4}) {}^{2} }{1 + ( \frac{2}{4}) {}^{2} } = \frac{1 - \frac{9}{16} }{4 + \frac{9}{16} } = \frac{7}{25}$

RHS

$= cos {}^{2} a - sin {}^{2} a = ( \frac{4}{5}) {}^{2} - ( \frac{3}{5} ) {}^{2} = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$

Since both the LHS & RHS

$= \frac{7}{25}$

RHS=LHS

RHS=LHShence,

$\frac{1 - tan {}^{2}a }{1 + tan {}^{2}a } = cos {}^{2} a - sin {}^{2}a$

is proved!

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Answer 2

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now, compare the LHS with the RHS

LHS

=1-tan²a/1+tan²a=1-(2/4)²/1+(2/4)²

=1-9/16 /4+9/16=7/25

since , both LHS & RHS = 7/25

RHS = LHS hence,

1-tan²a/1+tan²a = cos²a-sin²a

is proved!!!

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