Math, asked by amreen45, 1 year ago

Cos2 theta + tan2 theta -1/ sin2 theta = tan2 theta

Answers

Answered by vishwa9775
20

Answer:

Tan

tan ^{2}  \: theta

Step-by-step explanation:

For step by step explaination refer to the attachment.........

Attachments:
Answered by TRISHNADEVI
21

 \huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \:   \: QUESTION\:  \: } \mid}}}}}

 \sf{Prove  \:  \: that, } \\  \\  \huge{ \sf{\frac{cos {}^{2}A + tan {}^{2}A - 1  }{sin {}^{2}A } = tan {}^{2}A } }

 \huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \:   \: SOLUTION\:  \: } \mid}}}}}

  \tt{ L.H.S. = \frac{cos {}^{2}A + tan {}^{2}A - 1  }{sin {}^{2}A }}  \\  \\    \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{=  \frac{cos {}^{2}A + \frac{sin {}^{2} A}{cos {}^{2} A}   - 1  }{sin {}^{2}A }}   \\  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{=    \frac{cos {}^{2}A + \frac{sin {}^{2} A}{cos {}^{2} A}   - (sin {}^{2}A + cos {}^{2}A  )  }{sin {}^{2}A } }

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{= \frac{ \cancel{cos {}^{2}A} + \frac{sin {}^{2} A}{cos {}^{2} A}   - sin {}^{2}A  -  \cancel{ cos {}^{2}A  }  }{sin {}^{2}A }} \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{= \frac{ \frac{sin {}^{2} A - sin {}^{2}A.cos {}^{2}  A}{cos {}^{2} A} }{sin {}^{2}A }  }

\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{   = \frac{sin {}^{2} A - sin {}^{2} A.cos {}^{2} A}{cos {}^{2}A }  \times  \frac{1}{sin {}^{2}A }}  \\  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{ = \frac{sin {}^{2} A - sin {}^{2} A.cos {}^{2} A}{cos {}^{2}A .sin {}^{2} A} }\\  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{=   \frac{sin {}^{2}A \: (1 -cos {}^{2}A  ) }{cos {}^{2}A.sin {}^{2}A  }}

\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{= \frac{sin {}^{2} A \: .  \: \cancel{sin {}^{2} A}}{cos {}^{2}A \: . \:  \cancel{sin {}^{2} A } }} \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{= \frac{sin {}^{2} A}{cos {}^{2} A}}  \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{= tan {}^{2} A }\\  \\\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \tt{  = R.H.S.}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \text{ \underline{ \: Hence,  proved.  }}

 \underline{ \text{ \: FORMULA  \:  \:  USED :-  \: }}  \\  \\  \\  \bold{1. \:  \frac{sin  \: \theta}{cos \:  \theta}  = tan \:  \theta }\\ \\   \\  \bold{2. \: sin {}^{2}  \theta + cos {}^{2}   \theta = 1}

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