prove : tan square theta - sin square theta = tan square theta sin square theta
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Answered by
0
tan²-sin²=tan²sin²
sin² sin²
---- -
cos²
sin²-sin²cos²
-----------------
cos²
sin(1-cos²)
--------------
cos²
sin²*sin²
-----------
cos²
sin * sin
---
cos
tan²*sin²
hope it helps
sin² sin²
---- -
cos²
sin²-sin²cos²
-----------------
cos²
sin(1-cos²)
--------------
cos²
sin²*sin²
-----------
cos²
sin * sin
---
cos
tan²*sin²
hope it helps
Answered by
0
\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.
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