prove the following trignometry identity. tan squar -sin squar = tan squar×sin squar
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Answered by
0
Hey !!!
from LHS
tan² - sin²
=> sin²/cos²A - sin²A
=> sin² ( 1/cos² - 1 )
=> sin² ( sec²A - 1 )
=> sin² × tan² {•°• sec² - 1 = tan² }
=> tan² × sin² .
hence, RHS = LHS prooved
_____________________
Hope it helps you !!
@Rajukumar111
from LHS
tan² - sin²
=> sin²/cos²A - sin²A
=> sin² ( 1/cos² - 1 )
=> sin² ( sec²A - 1 )
=> sin² × tan² {•°• sec² - 1 = tan² }
=> tan² × sin² .
hence, RHS = LHS prooved
_____________________
Hope it helps you !!
@Rajukumar111
Answered by
1
\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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