Question

If x=sin^2theta and y= cos^2 theta is a solution of x+y-k^2=0 then find the value of k

Answer 1

Question :

If $x = \sin {}^{2} \beta$ and

$y = \cos {}^{2} ( \beta )$ is a solution of x+y-k^2=0 then find the value of k .

Some Trignometric Formulas:

• sin²A + cos²A = 1
• sec²A - tan²A = 1
• cosec²A - cot²A = 1
• sin2A = 2 sinA cosA
• cos2A = cos²A - sin²A

Solution :

given :

$x = \sin {}^{2} \beta$

$y = \cos {}^{2} ( \beta )$

we have to find the value of k ,if :

$x + y - k {}^{2} = 0$

⇒x + y = k²

Now put the values of x and y

$\sin {}^{2} ( \beta ) + \cos {}^{2} ( \beta ) = k {}^{2}$

we know that

sin²A + cos²A = 1

⇒k² = 1

${\purple{\boxed{\large{\bold{ k = \frac{ + }{} 1}}}}}$

Answer 2

Answer:

$\: \: \: \tt \pm1.$

Explanation :

$\tt \: \: \: \: \: \: x = { \sin }^{2} \theta. \\ \: \: \: \: \: \: \tt y = { \cos }^{2} \theta.$

According to equation,

$\: \: \: \: \: \: \leadsto \tt{x + y - {k}^{2} = 0.}$

$\: \: \: \: \: \: \leadsto \tt { \sin }^{2} \theta + {\cos}^{2} \theta - {k}^{2} = 0.$

$\: \: \: \: \: \: \tt \leadsto 1 - {k}^{2} = 0.$

$\: \: \: \: \: \: \tt \leadsto \cancel - {k}^{2} = \cancel- 1.$

$\tt \: \: \: \: \: \: \leadsto {k}^{2} = 1.$

$\tt \: \: \: \: \: \: \leadsto k = \sqrt{1} .$

$\tt \: \: \: \: \: \: \leadsto k = \pm1.$

Therefore, the value of k be 1.

Some information :

$\boxed{</p><p></p><p>\begin{minipage}{7 cm}</p><p></p><p>Fundamental Trigonometric Identities \\ \\</p><p></p><p> \sin^2\theta + \cos^2\theta=1 \\ \\</p><p></p><p>1+\tan^2\theta = \sec^2\theta \\ \\</p><p></p><p>1+\cot^2\theta = \text{cosec}^2 \, \theta </p><p></p><p>\end{minipage}</p><p></p><p>}$