Question

if 3 cot A = 4 , check whether 1 - Tan2 A / 1 + Tan2 A = cos2 A - sin2 A or not​

Answer 1

Given :-

• If $\bf{3cot A = 4}$ , check whether $\bf{\dfrac{1 -Tan^{2}A}{1 +Tan^{2}A} = Cos^{2}A - Sin^{2}A}$ or not.

Find :-

$\longrightarrow \sf{Prove :}$

$:\implies \bf{\dfrac{1 -Tan^{2}A}{1 +Tan^{2}A} = Cos^{2}A - Sin^{2}A}$

Solution -

As we know ,

• $:\implies \bf{CotA = \dfrac{Base}{Perpendicular}}$

• $\longmapsto{3cot A = 4}$
• $\longmapsto{cot A = \dfrac{4}{3}}$

Let we consider sides be in ratio " k "

$\longmapsto{cot A = \dfrac{4k}{3k}}$

Where,

• Base → 4k

• Perpendicular → 3k

For Hypotenuse :

By using Pythagoras Theorem -

$:\implies \bf{Hypotenuse^{2} = Base^{2} + Perpendicular^{2}}$

$:\implies \bf{AC^{2} = AB^{2} + BC^{2}}$

$:\implies \bf{AC^{2} = (4k)^{2} + (3k)^{2}}$

$:\implies \bf{AC^{2} = 16k^{2} + 9k^{2}}$

$:\implies \bf{AC^{2} = 25k^{2}}$

$:\implies \bf{AC = 5k}$

Let's find out other Ratios now,

⠀⠀⠀⠀⠀For sin A :

$:\implies \bf{SinA = \dfrac{Perpendicular}{Hypotenuse}}$

$:\implies \sf{SinA = \dfrac{BC}{AC}}$

$:\implies \sf{SinA = \dfrac{3k}{5k}}$

$:\implies \bf{SinA = \dfrac{3}{5}}$

⠀⠀⠀⠀⠀For Cos A :

$:\implies \bf{CosA = \dfrac{Base}{Hypotenuse}}$

$:\implies \sf{CosA = \dfrac{AB}{AC}}$

$:\implies \sf{CosA = \dfrac{4k}{5k}}$

$:\implies \bf{CosA = \dfrac{4}{5}}$

⠀⠀⠀⠀⠀For Tan A :

$:\implies \bf{TanA = \dfrac{Perpendicular}{Base}}$

$:\implies \sf{TanA = \dfrac{BC}{AB}}$

$:\implies \sf{TanA = \dfrac{3k}{4k}}$

$:\implies \bf{TanA = \dfrac{3}{4}}$

Proof :

For LHS ➤

We know :-

• $:\implies \bf{CotA = \dfrac{Base}{Perpendicular}}$
• $:\implies \bf{SinA = \dfrac{3}{5}}$
• $:\implies \bf{CosA = \dfrac{4}{5}}$
• $:\implies \bf{TanA = \dfrac{3}{4}}$

Then, Substituting the respective values , we get :

$:\implies \bf{\dfrac{1-tan^{2}A}{1+tan{2}^A}}$

$:\implies \bf{\bigg (1 - \dfrac{3}{4}\bigg)^{2} \div \bigg (1 + \dfrac{3}{4}\bigg)^{2}}$

$:\implies \bf{\bigg (\dfrac{16 - 9}{16}\bigg) \div \bigg(\dfrac{16 + 9}{16}\bigg)}$

$:\implies \sf{ \dfrac{7}{16} \div \dfrac{25}{16}}$

By reciprocal -

$:\implies \sf{ \dfrac{7}{16} \times \dfrac{16}{25}}$

$:\implies \bf{ \dfrac{7}{25}}$

And RHS :-

Again, Putting the Values ,

$:\implies \bf{Cos^{2}A - Sin^{2}A}$

$:\implies \bf{\bigg(\dfrac{4}{5}\bigg)^{2} - \bigg(\dfrac{3}{5}\bigg)^{2}}$

$:\implies \sf{ \dfrac{16}{25} - \dfrac{9}{25}}$

$:\implies \bf{ \dfrac{7}{25}}$

$:\implies \sf{ \dfrac{7}{25} = \dfrac{7}{25}}$

L.H.S = R.HS

Hence,

Proved !!

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