if sinA+cosA=root 2 cosA then find the value of cotA.pls answer fast as tomorrow is my maths exam
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sin A + cos A =√cos A
then, sin A=√cos A - cos A
sin A=√cos A(1-√(cosA))......(i)
As we know,
cot A = cos A / sin A
cot A = cosA/√(cosA)(1-√(cosA))
cot A =√(cos A)/(1-√(cos A))
cot A =√(cos A)/(1-√(cos A)) x (1+√(cos A)/1+√(cos A)) (By taking conjugate)
cot A =√(cosA)(1+√(cosA))/1-cosA
cot A =[√(cosA) + cosA/1-cos A] x (cos A+ 1/cos A + 1) (again taking conjugate)
cot A =[{√(cosA) + cosA}(cosA+1)}/1-cos^2A]
∴cot A =(cosA√(cosA)+√(cosA)+cos^2A+cosA)/sin^2A
Hope it Helps..!
sin A + cos A =√cos A
then, sin A=√cos A - cos A
sin A=√cos A(1-√(cosA))......(i)
As we know,
cot A = cos A / sin A
cot A = cosA/√(cosA)(1-√(cosA))
cot A =√(cos A)/(1-√(cos A))
cot A =√(cos A)/(1-√(cos A)) x (1+√(cos A)/1+√(cos A)) (By taking conjugate)
cot A =√(cosA)(1+√(cosA))/1-cosA
cot A =[√(cosA) + cosA/1-cos A] x (cos A+ 1/cos A + 1) (again taking conjugate)
cot A =[{√(cosA) + cosA}(cosA+1)}/1-cos^2A]
∴cot A =(cosA√(cosA)+√(cosA)+cos^2A+cosA)/sin^2A
Hope it Helps..!
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