Question

if 3 cot A = 4cheack Whether 1-tan² A/1+tan²A =cos²A
-sin²A or not ​

Answer 1

Answer:

Formula used:

Tan A = sinA/cosA

Cot A= cosA / sinA

Sin²A+ cos²A= 1

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Answer:

1+tan²A/1+cot²A = tan²A

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Hope this will help you...

Answer 2

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Given :-

$3 \cot(A) = 4$

$↪ \cot(A) = \frac{4}{3}↪ \frac{b}{p} = \frac{4}{3}$

Let:-

$= B = 4k \: \: \: and \: P = 3k$

Where , K is any positve integer .

Draw a right angled ΔABC right angled at B .

[ Note :-Figure in Attachment ]

In right angled ΔABC , H²= P²+B² [by pytha. ]

${H}^{2} = { (3k)}^{2} + {(4k)}^{2}$

$◻ {9k}^{2} + {16k}^{2} = {25k}^{2}$

$◻ {h}^{2} = 5k$

Therefore :-

$↪\tan(A) = \frac{1}{ \cot(A) } = \frac{3}{4}$

$↪\cos(A) = \frac{b}{h} = \frac{4k}{5k} = \frac{4}{5}$

$↪\sin(A) = \frac{p}{h} = \frac{3k}{5k} = \frac{3}{5}$

Now LHS :-

$↪\frac{1 - \tan ^{2} (A) }{1 + \tan ^{2} (A) } = \frac{ { 1 - (\frac{3}{4}) }^{2} }{ {1 + ( \frac{3}{4} )}^{2} }$

[Since , Tan A = 3/4 ]

$↪\frac{1 - \frac{9}{16} }{1 + \frac{9}{16} } = \frac{ \frac{16 - 9}{16} }{ \frac{16 + 9}{16} } = \frac{7}{5} ......(1eq)$

RHS = Cos²A - Sin²A

$↪( \frac{4}{5})^{2} - ( { \frac{3}{5} )}^{2} = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} .......(2eq)$

From 1st and 2nd equation

LHS = RHS

$↪ \frac{7}{25} = \frac{7}{25}$

✏Hence Proved