Question

solve the given question​

Answer 1

Given,

$\longrightarrow\csc\theta-\sin\theta=a^3$

$\longrightarrow\dfrac{1}{\sin\theta}-\sin\theta=a^3$

$\longrightarrow\dfrac{1-\sin^2\theta}{\sin\theta}=a^3$

$\longrightarrow\dfrac{\cos^2\theta}{\sin\theta}=a^3$

$\longrightarrow\cos\theta=\sqrt{a^3\sin\theta}$

Or,

$\longrightarrow\cos\theta=a^{\frac{3}{2}}\sin^{\frac{1}{2}}\theta\quad\quad\dots(1)$

Given,

$\longrightarrow\sec\theta-\cos\theta=b^3$

$\longrightarrow\dfrac{1}{\cos\theta}-\cos\theta=b^3$

$\longrightarrow\dfrac{1-\cos^2\theta}{\cos\theta}=b^3$

$\longrightarrow\dfrac{\sin^2\theta}{\cos\theta}=b^3$

From (1),

$\longrightarrow\dfrac{\sin^2\theta}{a^{\frac{3}{2}}\sin^{\frac{1}{2}}\theta}=b^3$

$\longrightarrow\dfrac{\sin^2\theta}{\sin^{\frac{1}{2}}\theta}=a^{\frac{3}{2}}b^3$

$\longrightarrow\sin^{\frac{3}{2}}\theta=\left(ab^2\right)^{\frac{3}{2}}$

$\longrightarrow\sin\theta=ab^2\quad\quad\dots(2)$

Putting (2) in (1),

$\longrightarrow\cos\theta=a^{\frac{3}{2}}(ab^2)^{\frac{1}{2}}$

$\longrightarrow\cos\theta=a^{\frac{3}{2}}a^{\frac{1}{2}}b$

$\longrightarrow\cos\theta=a^2b\quad\quad\dots(3)$

Now, we know that,

$\longrightarrow \sin^2\theta+\cos^2\theta=1$

From (2) and (3),

$\longrightarrow (a^2b)^2+(ab^2)^2=1$

$\longrightarrow a^4b^2+a^2b^4=1$

$\longrightarrow\underline{\underline{a^2b^2(a^2+b^2)=1}}$

Hence Proved!

Answer 2

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.