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Prove that sin thita -cos thita …

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Prove that sin thita -cos thita +1/sin thita + cos thita - 1 =1/sec thita + tan thita , using identity sec² thita =1+ tan²thita

Posted by Arya Jadav 5 years, 5 months ago

CBSE > Class 10 > Mathematics

4 answers

Arya Jadav 5 years, 5 months ago

Ok Thanks

Pankaj Malik 5 years, 5 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

Pankaj Malik 5 years, 5 months ago

Its mistake sorry

Pankaj Malik 5 years, 5 months ago

In the question, We have been provided a trigonometric function, Therefore, on simplifying the given equation by putting in the values of the respective terms in the LHS, we get, On dividing by 'cosθ' in both Numerator and Denominator, we get, Now, we know that, sec²θ - tan²θ = 1 On putting in the given equation, we get, i.e. Now, we can see here that, LHS of the equation is equal to the RHS. Therefore, we can see that the equation has been proved. Hence, Proved.

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