Question

x sin^3 theta + y cos^3 theta = sintheta costheta and x sin theta = ycostheta prove that
x^2 + y² = 1.​

Answer 1

Solution

• Given

$\implies \mathsf{x\: { \sin(\theta) }^{3} + y\: {\cos(\theta) }^{3} \: = \sin(\theta) \cos(\theta) }$

$\implies \mathsf{x \sin(\theta)\: = y \cos(\theta)}$

• To prove

$\longrightarrow \mathsf{x^2 + y^2 = 1}$

• Proof

*To solve such questions first we will try to break LHS (As per your easiness) to a simpler form and convert it into a useful form.

Therefore Taking

$\large \bold{\underline{\underline{LHS \implies}}}$

$\longrightarrow \mathsf{x \: { \sin(\theta) }^{3} + y \: {\cos(\theta) }^{3}}$

$\longrightarrow \mathsf{x \: { \sin(\theta) }^{2} \sin( \theta) + y \: {\cos(\theta) }^{2} \cos( \theta) }$

Now we know that,

$\implies \mathsf{x \sin(\theta)\: = y \cos(\theta)}$

therefore putting this value in equation we have,

$\longrightarrow \mathsf{y \cos( \theta) \: { \sin(\theta) }^{2} + y \: {\cos(\theta) }^{2} \cos( \theta) }$

Taking y cos(0) common we have,

$\longrightarrow \mathsf{y \cos( \theta) \left( { \sin(\theta) }^{2} + {\cos(\theta) }^{2}\right)}$

Now By formula,

$\boxed{\boxed{ \bigstar \: \mathsf{ { \sin( \theta) }^{2} + { \cos( \theta) }^{2} \: = 1 }}}$

We have,

$\longrightarrow \mathsf{y \cos( \theta) = \sin( \theta) \cos( \theta) }$

Now taking Cos(theta) to RHS

$\longrightarrow \mathsf{y \cos( \theta) = \sin( \theta) \cos( \theta) }$

$\longrightarrow \mathsf{y = \dfrac{\sin( \theta) \cos( \theta)}{ \cos( \theta) } }$

$\longrightarrow \mathsf{y = \sin( \theta) }$

Therefore we have,

$\implies \mathsf{y = \sin( \theta) }$

and

$\implies \mathsf{x = \cos( \theta) }$

Therefore now putting this value we have

$\longrightarrow \mathsf{x^2 + y^2 = 1}$

$\longrightarrow \mathsf{{\cos( \theta)}^{2} + {\sin( \theta)}^{2} = 1}$

Now from the above formula we have

$\longrightarrow \mathsf{1 = 1}$

Thus we have,

• LHS = RHS

• Hence proved

Answer 2

Question :------- we have to prove x²+y² = 1

Given :------

• xsin³@ + y cos³@ = sin@ × cos@
• x sin @ = y cos@

Formula to be used :---

• sin²@ + cos²@ = 1

Solution :--------

let

xsin³@ + y cos³@ = sin@ × cos@

(x sin@)sin²@ + (ycos@)cos²@ = sin@ × cos@

since x sin@ = y cos@

putting value we get,

(ycos@)sin²@ + (ycos@)cos²@ = sin@ × cos@

Taking (ycos@) common now ,

(ycos@)[sin²@+cos²@] = sin@ × cos@

→ ycos@ = sin@ × cos@

→ y = sin@ -------------------- Equation (1)

______________________________

Now, Again putting value of (ycos@) this time we get,

(x sin@)sin²@ + (ycos@)cos²@ = sin@ × cos@

→ (x sin@)sin²@ + (xsin@)cos²@ = sin@ × cos@

Taking (x sin@) common Now,

(xsin@)(sin²@+cos²@) = sin@ × cos@

→ x sin@ = sin@ × cos@

→ x = cos@ ------------------------ Equation (2)

______________________________

Now , Squaring both Equation and adding them we get,

x² + y² = sin²@ + cos²@

x²+y² = 1 (Proved)

(Hope it helps you)

______________________________

$\color {red}\large\bold\star\underline\mathcal{Extra\:Brainly\:Knowledge:-}$

Trignometric identity :---

→ sin²θ = 1 − cos²θ

→ cos²θ = 1 − sin²θ

→ tan²θ + 1 = sec²θ

→ tan²θ = sec²θ − 1

→ cot²θ + 1 = csc²θ

→ cot²θ = csc²θ − 1

$\mathcal{BE\:\:BRAINLY}$