Question

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

Answer 1

Answer:

We have to prove that.......

1–tan²(45°–A)/1+tan(45°–A) = cosec2A

On taking L.H.S......

1–tan²(45°–A)/1+tan(45°–A)...........(1)

As we know that.....sec²A = 1+tan²A then from equation (1)

=>1–tan²(45°–A)/sec²(45°–A)

=>1/sec² (45°–A) – tan²(45–A)/sec²(45–A)

=>cos²(45–A) – sin²(45–A)

{ Because cosA = 1/secA and tanA/secA = sinA}

=>cos2(45°–A)

{ Because 1 – cos²A = sin2A }

=>sin(90°–2A)

=>cosec2A { Because sin(90–A) = cosecA }

= R.H.S.

Answer 2

$\huge\boxed{\fcolorbox{black}{pink}{answer}}$

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