Question

if $\sf tan\: \theta =\dfrac{a}{b}$
prove that,
$\tt \bigg( \dfrac{sin^2 \: \theta - cos^2\: \theta}{sin^2 \: \theta + cos^2\: \theta} \bigg) = \bigg( \dfrac{a^2-b^2}{a^2+b^2} \bigg)$
​

Answer 1

Here is your answer ❤️⏬

We already learnt that,

$\sf tan \theta = \frac{Sin \theta }{Cos \theta}$

So, According to the question we can say that,

$\sf tan \theta = \frac{a}{b} = \frac{Sin \theta}{Cos \theta }$

Hence, $\sf Sin \theta = a$ and $\sf Cos \theta = b$

Now, we have to put these values in $\sf \bigg( \frac{sin^2 \theta - cos^2 \theta}{sin^2 \theta + cos^2 \theta} \bigg)$

$\sf \bigg( \frac{sin^2 \theta - cos^2 \theta}{sin^2 \: \theta + cos^2\: \theta} \bigg)$

$\sf \bigg( \frac{(a)^{2} - (b)^{2}}{(a)^{2} + (b)^{2}} \bigg)$

Hence It's Proof

$\sf \bigg( \frac{a^{2} - b^{2}}{a^{2} + b^{2}} \bigg)$

More information about Trigonometry :

• $\sf Tan \theta = \frac{Sin \theta }{Cos \theta }$

• $\sf Cot \theta = \frac{1}{Tan \theta } = \frac{Cos \theta }{Sin \theta}$

• $\sf Sin² \theta + Cos² \theta = 1$

• $\sf Sec² \theta - Tan² \theta = 1$

• $\sf cosec² \theta = 1 + cot² \theta$

I hope it helps you ❤️✔️

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:tan \theta \: = \: \dfrac{a}{b}$

Now, Consider

$\rm :\longmapsto\:\dfrac{ {sin}^{2} \theta - {cos}^{2} \theta }{ {sin}^{2} \theta + {cos}^{2} \theta }$

$\red{ \sf{ \: On \: dividing \: numerator \: and \: denominator \: by \: {cos}^{2}\theta }}$

$\rm \:  =  \: \dfrac{\dfrac{ {sin}^{2} \theta - {cos}^{2} \theta }{ {cos}^{2} \theta } }{\dfrac{ {sin}^{2} \theta + {cos}^{2} \theta }{ {cos}^{2} \theta } }$

$\rm \:  =  \: \dfrac{\dfrac{ {sin}^{2} \theta }{ {cos}^{2} \theta } - 1}{\dfrac{ {sin}^{2} \theta}{ {cos}^{2} \theta } + 1}$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \frac{sinx}{cosx} = tanx \: }}}$

So, using this, we get

$\rm \:  =  \: \dfrac{ {tan}^{2} \theta - 1}{ {tan}^{2} \theta + 1}$

$\rm \:  =  \: \dfrac{\dfrac{ {a}^{2} }{ {b}^{2}} - 1}{\dfrac{ {a}^{2} }{ {b}^{2} } + 1}$

$\rm \:  =  \: \dfrac{\dfrac{ {a}^{2} - {b}^{2} }{ {b}^{2}}}{\dfrac{ {a}^{2} + {b}^{2} }{ {b}^{2} }}$

$\rm \:  =  \: \dfrac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} }$

Hence, Proved

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1