Question

prove : tan square theta - sin square theta = tan square theta sin square theta

Answer 1

tan²-sin²=tan²sin²
sin²      sin²
----    -
cos²
sin²-sin²cos²
-----------------
cos²
sin(1-cos²)
--------------
cos²
sin²*sin²
-----------
cos²
sin  * sin
---
cos
tan²*sin²
hope it helps

Answer 2

\sin^2A\cos^2B-\cos^2A\sin^2B=\sin^2A-\sin^2B proved.

Step-by-step explanation:

Consider the provided information.

\sin^2A\cos^2B-\cos^2A\sin^2B=\sin^2A-\sin^2B

Consider the LHS.

\sin^2A\cos^2B-\cos^2A\sin^2B

\sin^2A(1-\sin^2B)-(1-\sin^2A)\sin^2B               (∴\cos^2x=1-\sin^2x)

\sin^2A-\sin^2A\sin^2B-\sin^2B+\sin^2A\sin^2B

\sin^2A-\sin^2B

Hence, proved.