tan/(1 --cot 0)+coto/(1 - tan ) = (1 + sec 0 cosec 0)
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Step-by-step explanation:
Given Tan/(1 --cot 0) + coto/(1 - tan ) = (1 + sec 0 cosec 0)
- Consider the left hand side we have
- tan theta / 1 – cot theta + cot theta / 1 – tan theta
- tan theta / 1 – 1/tan theta + 1/tan theta / 1 – tan theta
- tan theta / tan theta – 1 / tan theta + 1 / tan theta (1 – tan theta)
- tan^2 theta / tan theta – 1 + 1 / tan theta (1 – tan theta)
- -tan^2 theta / 1 – tan theta + 1 / tan theta (1 – tan theta) (taking tan theta (1 – tan theta as lcm we get)
- -tan^3 theta + 1 / tan theta (1 – tan theta)
- 1^3 – tan^3 theta / tan theta (1 – tan theta)
- So a^3 – b^3 = (a – b)(a^2 + ab + b^2)
- (1 – tan theta) (1 + tan theta + tan^2 theta) / tan theta (1 – tan theta)
- 1 + tan theta + tan^2 theta / tan theta
- 1 + tan^2 theta + tan theta / tan theta
- sec^2 theta + tan theta / tan theta
- sec^2 theta / tan theta + tan theta / tan theta (by splitting)
- sec^2 theta cot theta + 1
- 1/cos^2 theta x cos theta / sin theta + 1
- 1 / cos theta sin theta + 1
- sec theta cosec theta + 1
- Or 1 + sec theta cosec theta (r.h.s)
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https://brainly.in/question/1236608
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