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आप 0 -0089+1 न्म्न्ग्ञः=Example 15 : Pro"ve 08 - == "/J o sin+cos~=1 sec-tan1+ tan® 6. i sec’ 0 |
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Answer» (sinθ - cosθ +1 )/(sinθ +cosθ -1) dividing numerator and denominator by cosθ [(sinθ - cosθ +1 )cosθ]/[(sinθ +cosθ -1)/cosθ]=(tanθ -1 + secθ )/(tanθ +1 - secθ)=(tanθ + secθ -1)/(tanθ - secθ+1) As, sec²θ- tan²θ = 1(secθ -tanθ)(secθ +tanθ) = 1 putting this in numerator,[(tanθ + secθ -(sec²θ- tan²θ)]/(tanθ - secθ+1)=[(tanθ + secθ) -(secθ- tanθ)(secθ+tanθ)]/(tanθ - secθ+1)=(tanθ+secθ)[1- (secθ - tanθ)]/(tanθ - secθ+1)=(tanθ+secθ)[1- secθ + tanθ)]/(tanθ - secθ+1)=(tanθ+secθ) Now, multiplying and dividing by(secθ- tanθ)[(tanθ+secθ)×(secθ- tanθ)]/(secθ- tanθ)=(sec²θ- tan²θ)/(secθ- tanθ)= 1/(secθ- tanθ)=RHS Hence proved |
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