If tan theta=p/q show that pain theta-qcos theta/p sin theta+question theta =psquare-q square/p square+q square
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Let me tell you an interesting fact about Trigonometry."Triangle" > "Trigonometry"Remember some formulae now :sin²θ + cos²θ = 1sec²θ - tan²θ = 1cosec²θ - cot²θ = 1Want to learn more!Here it is :sin(A + B) = sinA cosB + cosA sinBsin(A - B) = sinA cosB - cosA sinBcos(A + B) = cosA cosB - sinA sinBcos(A - B) = cosA cosB + sinA sinB(SEE THE ATTACHMENT).........
hope it's helpful..............
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