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SINA-COSA+1/SINA+COS-1=1/SECA-TANA

Posted by Archana Yadav 6 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

#Miss_Raagini ???? 6 years, 9 months ago

LHS = (sinθ - cosθ + 1)/(sinθ + cosθ - 1) dividing by cosθ both Numerator and denominator = (sinθ/cosθ - cosθ/cosθ + 1/cosθ)/(sinθ/cosθ + cosθ/cosθ - 1/cosθ) = (tanθ + secθ - 1)/(tanθ - secθ + 1) Multiply (tanθ - secθ) with both Numerator and denominator = (tanθ + secθ - 1)(tanθ - secθ)/(tanθ - secθ + 1)(tanθ - secθ) = {(tan²θ - sec²θ) - (tanθ - secθ)}/(tanθ - secθ + 1)(tanθ - secθ) = (-1 - tanθ + secθ)/(tanθ - secθ + 1)(tanθ - secθ) [ ∵ sec²x - tan²x = 1 ] = -1/(tanθ - secθ) = 1/(secθ - tanθ) = RHS Hope it helps archana ?

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Posted by Hari Anand 6 months, 3 weeks ago

CBSE > Class 10 > Mathematics

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Posted by Kanika . 1 month, 2 weeks ago

CBSE > Class 10 > Mathematics