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CosA +sinA = √2cosA

squaring

⇒(cosA + sin A)² = (√2cosA)²

⇒cos²A + sin²A + 2sinAcosA = 2cos²A

⇒1 - sin²A + 1 - cos²A + 2sinAcosA = 2cos²A

⇒2 - 2cos²A = cos²A + sin²A - 2sinAcosA

⇒2(1 - cos²A)= (cosA - sinA)²

⇒ cosA - sinA = √[2sin²A]

⇒cosA-sinA = √2sinA

hence proved

squaring

⇒(cosA + sin A)² = (√2cosA)²

⇒cos²A + sin²A + 2sinAcosA = 2cos²A

⇒1 - sin²A + 1 - cos²A + 2sinAcosA = 2cos²A

⇒2 - 2cos²A = cos²A + sin²A - 2sinAcosA

⇒2(1 - cos²A)= (cosA - sinA)²

⇒ cosA - sinA = √[2sin²A]

⇒cosA-sinA = √2sinA

hence proved

squaring on both sides we get

(cos A + sin A)² = (√2cos A)²

it is in the form of (a+b)^2

formula for (a+b)^2=a^2+ab+b^2

by comparing here we have a=cos A and b=sin A

cos²A + sin²A + 2sinAcosA = 2cos²A

we know that

sin^2A+cos^2 A=1

from this we can write

sin^2 A=1-cos^2 A

and cos^2 A=1-sin^2 A

1 - sin²A + 1 - cos²A + 2 sinAcosA = 2 cos²A

adding like terms and sin^2 A+cos^2 A -2 sinA cosA -2 cos^2 A on both sides

2 - 2 cos²A = cos²A + sin²A - 2 sinAcosA

taking 2 as common in left side of the equation

2(1 - cos²A)= (cos A - sin A)²

cos A - sin A = √[2 sin²A]

cos A-sin A = √2 sin A

hence it is proved

tanA = sinA/cosA

cotA = cosA/sinA

1 + cot^2A = cosec^2A

tan^2A + 1 = sec^2A

cosecA = 1/sinA

secA = 1/cosA

cotA = 1/tanA

(Only use the above identities to prove the question)