Question

can you answer this question ​

Answer 1

Answer:

LHS:-

tan²A-sin²A=sin²A/cos²A -sin²A

sin²A.sec²A-sin²A

take common sin²A

sin²A(sec²A-1)

sin²A.tan²A

tan²A.sin²A

proved

mark as Brainlist answer

Answer 2

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:

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

$\sf{tan^2\theta-sin^2\theta}\\\sf{\dfrac{sin^2\theta}{cos^2\theta}-sin^2\theta}\\\sf{\dfrac{sin^2\theta-sin^2\theta\:cos^2\theta}{cos^2\theta}}\\\sf{Taking\:sin^2\theta\:as\:common}\\\sf{\dfrac{sin^2\theta\left(1-cos^2\theta\right)}{cos^2\theta}}\\\sf{\dfrac{sin^2\theta}{cos^2\theta}sin^2\theta}\\\sf{\tan^2\theta\:sin^2\theta}$

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

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