Question

If tanA-cotA=a and cosA+sinA=b, then show that (a^2+4)(b^2-1)^2=4

Answer 1

(a²+4)(b²-1)²=4

Step-by-step explanation:

Given

tan A - cot A =a

⇒(tan A - cot A)² =a²    [ squaring both sides]

⇒tan²A+cot²A-2 tan A cot A= a²

⇒tan²A+cot²A-2 =a²    $[\because tan A cotA=tanA\frac{1}{tanA} =1]$

And

cos A+sin A= b

⇒(cos A+sin A)²= b²  [ squaring both sides]

⇒cos²A+sin²A+2sinAcosA= b²

⇒1+2sinAcosA= b²

L.H.S= (a²+4)(b²-1)²

         =(tan²A+cot²A-2+4)(1+2sinAcosA-1)²

        =(tan²A+cot²A+2)4sin²A cos²A

       =tan²A.4sin²A cos²A+cot²A.4sin²A cos²A+2.4sin²A cos²A

        $=\frac{sin^2A}{cos^2A} 4sin^2A cos^2A+\frac{cos^2A}{sin^2A} 4sin^2A cos^2A+8sin^2A cos^2A$

       =4 sin⁴A +4 cos⁴A+8 sin²A cos²A

       =4(sin⁴A + cos⁴A+2 sin²A cos²A)

        =4(sin²A+cos²A)²

        =4.1

        =4 = R.H.S (Proved)