Question

please solve ques 9 and 10.
IT'S URGENT

Answer 1

1.

Consider cosec theta - sin theta = a³

⇒ !/sin theta - sin theta = a³

⇒ 1 - sin² theta/sin theta = a³

cos² theta/ sin theta = a³ → (1)

⇒ (cos² theta/sin theta)²/³ = (a³)²/³

⇒ cos⁴/³ theta/sin²/³ theta = a² → (2)

Now consider, sec theta - cos theta = b³

⇒ 1/cos theta - cos theta = b³

⇒ 1 - cos²theta/cos theta = b³

⇒ sin² theta/cos theta = b³ → (3)

⇒ (sin² theta/cos theta)²/³ = (b³)²/³

⇒ sin⁴/³ theta/cos²/³ theta = b² → (4)

Multiply (2) and (4), we get

(cos⁴/³ theta/sin²/³ theta)× (sin⁴/³ theta/cos²/³ theta) = a²b² → (5)

a² + b² =(cos⁴/³ theta/sin²/³ theta) + (sin⁴/³ theta/cos²/³ theta)

(cos² theta + sin² theta)/(sin²/³ theta cos²/³ theta)

= 1/sin²/³ theta cos²/³ theta

Consider, a²b²(a²+b²) = (sin²/³ theta cos²/³ theta) × 1/sin²/³ theta cos²/³ theta

= 1 Hence proved.

2.

Given, LHS = root tan a tan b + tan a cot b/sin a sec b - sin^2 b/cos^2 a

      = root tan a tan(90-a) + tan a cot(90-a)/sin a sec(90-a) - sin^2(90-a)/cos^2 a

      = root tan a.cot a + tan a.tana/sin a.cosec a - cos^2 a/cos^2 a

      = root 1+tan^2 a/1 - 1

     = root tan^2 a

    = tan a.

ꃅꂦᖘꍟ ꀤ꓄ ꃅꍟ꒒ᖘꌗ ꌩꂦꀎ