Question

If cosec (tita) - sin(tita)= x^3 and sec(tita)- cosec(tita) = y^3. then prove that x^2× y^2 (x^2+y^2)= 1
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Answer 1

correct Question:- if $\large\rm { \cosec \theta - \sin \theta = x^{3}}$

$\large\rm { \sec \theta - \cosec \theta = y^{3}}$

then prove that $\large\rm { x^{2} y^{2} (x^{2} + y^{2} = 1}$

Solution:-

$\large\rm { \implies \frac{1}{\sin \theta} - \sin \theta = x^{3}}$

$\large\rm { \implies \frac{ \cos^{2} \theta}{ \sin \theta} = x^{3}}$

$\large\rm { \leadsto x = ( \frac { \cos^{2} \theta}{\sin \theta} ) ^{\frac{1}{3}}}$

Similarly, $\large\rm { \leadsto y = ( \frac { \sin^{2} \theta}{\cos \theta} ) ^{\frac{1}{3}}}$

$\large\rm { \therefore x^{2} y^{2} ( x^{2} + b^{2} ) = x^{4} y^{2} + x^{2} y^{4}}$

Substituting the values,

$\large\rm { \cos^{2} \theta + \sin^{2} \theta = 1}$

$\large\boxed{\rm { \therefore x^{2} y^{2} ( x^{2} + y^{2} ) = 1}}$

Answer 2

Answer:

= x^{3}}⟹

sinθ

1

−sinθ=x

3

cos^{2}sin theta = x^{3}}⟹

sinθ

cos

2

θ

=x

3

\large\rm { \leadsto x = ( \frac { \cos^{2} \theta}{\sin \theta} ) ^{\frac{1}{3}}}⇝x=(

sinθ

cos

2

θ

)

3

1

Similarly, \large\rm { \leadsto y = ( \frac { \sin^{2} \theta}{\cos \theta} ) ^{\frac{1}{3}}}⇝y=(

cosθ

sin

2

θ

)

3

1

\large\rm { \therefore x^{2} y^{2} ( x^{2} + b^{2} ) = x^{4} y^{2} + x^{2} y^{4}}∴x

2

y

2

(x

2

+b

2

)=x

4

y

2

+x

2

y

4

Substituting the values,

\large\rm { \cos^{2} \theta + \sin^{2} \theta = 1}cos

2

θ+sin

2

θ=1

\large\boxed{\rm { \therefore x^{2} y^{2} ( x^{2} + y^{2} ) = 1}}

∴x

2

y

2

(x

2

+y

2

)=1