Question

pozzzzz Solve fast. ​

Answer 1

✮ Question ✮

$\sf{Prove\:that\:tan^2\theta-sin^2\theta=tan^2\theta\:sin^2\theta}$

✮ Identities Used ✮

$\sf{1-cos^2\theta=sin^2\theta}\\\sf{\dfrac{sin\theta}{cos\theta}=tan\theta}$

✮ Solution ✮

$\sf{Taking\:L.H.S,}$

$\sf{tan^2\theta-sin^2\theta}$

$\sf{\dfrac{sin^2\theta}{cos^2\theta}-sin^2\theta}$

$\sf{\dfrac{sin^2\theta-sin^2\theta\:cos^2\theta}{cos^2\theta}}$

$\sf{Taking\:sin^2\theta\:as\:common}$

$\sf{\dfrac{sin^2\theta\left(1-cos^2\theta\right)}{cos^2\theta}}$

$\sf{\dfrac{sin^2\theta}{cos^2\theta}sin^2\theta}$

✮ $\boxed{\sf{\tan^2\theta\:sin^2\theta}}$

$\sf{L.H.S=R.H.S}$ ✮

✮ $\sf{\mathrm{HENCE\:PROVED}}$ ✮

Answer 2

$\huge\underbrace\mathrm\purple{Answer}$

$\sf{Taking\:L.H.S,}$

$\sf{tan^2\theta-sin^2\theta}$

$\sf{\dfrac{sin^2\theta}{cos^2\theta}-sin^2\theta}$

$\sf{\dfrac{sin^2\theta-sin^2\theta\:cos^2\theta}{cos^2\theta}}$

$\sf{Taking\:sin^2\theta\:as\:common}$

$\sf{\dfrac{sin^2\theta\left(1-cos^2\theta\right)}{cos^2\theta}}$

$\sf{\dfrac{sin^2\theta}{cos^2\theta}sin^2\theta}$

$\boxed{\sf{\tan^2\theta\:sin^2\theta}}$

$\sf{L.H.S=R.H.S}$

$\sf{\mathrm{HENCE\:PROVED}}$

ʜᴏᴘᴇ ʏᴏᴜ ғɪɴᴅ ɪᴛ ʜᴇʟᴘғᴜʟ

ᴛʜᴀɴᴋ ʏᴏᴜ