Question

If x = cosec(θ) - sin(θ) and, y = sec(θ) - cos(θ) then find the Value of :
x²y²(x² + y² + 3)

- SSC CGL Tier I​

Answer 1

Answer:

1

Step-by-step explanation:

x = cosec∅ - sin∅ and y = sec∅ - cos∅

We have to find the value of x²y²(x² + y² + 3)

=> (cosec∅ - sin∅)²(sec∅ - cos∅)²[(cosec∅ - sin∅)² + (sec∅ - cos∅)² + 3]

=> (cosec∅ - sin∅)²(sec∅ - cos∅)²[cosec²∅ + sin²∅ - 2cosecsin∅ + sec²∅ + cos²∅ - 2sec∅cos∅ + 3]

=> (cosec∅ - sin∅)²(sec∅ - cos∅)²[cosec²∅ + 1 - 2 + sec²∅ - 2 + 3]

=> (cosec∅ - sin∅)²(sec∅ - cos∅)²[cosec²∅ + sec²∅]

=> (1/sin∅ - sin∅)²(1/cos∅ - cos∅)²[1/sin²∅ + 1/cos²∅]

=> (1 - sin²∅/sin∅)²(1 - cos²∅/cos∅)²[sin²∅ + cos²∅/sin²∅cos²∅]

=> (cos²∅/sin∅)²(sin²∅/cos∅)²[1/sin²∅cos²∅]

=> (cos⁴∅/sin⁴∅)(sin²∅/cos²∅)(1/sin²∅cos²∅)

=> (sin²∅cos²∅/sin²∅cos²∅)

=> 1

Hence, the value of x²y²(x² + y² + 3) = 1.

#Hope my answer helped you!

Answer 2

AnswEr :

$\bold{Given} \begin{cases} \sf{x= \csc( \theta) - \sin( \theta) } \\ \sf{y=} \sec( \theta) - \cos( \theta) \\ \sf{ To \: Find :{x}^{2} {y}^{2}( {x}^{2} + {y}^{2} + 3) }\end{cases}$

⇝ x = cosec(θ) - sin(θ)

⇝ x = 1 / sin(θ) - sin(θ)

⇝ x = (1 - sin²(θ)) / sin(θ)

⇝ x = cos²(θ) / sin(θ) —eq. ( I )

⠀

⇝ y = sec(θ) - cos(θ)

⇝ y = 1 / cos(θ) - cos(θ)

⇝ y = (1 - cos²(θ)) / cos(θ)

⇝ y = sin²(θ) / cos(θ) —eq. ( II )

• Now let's Head to the Question :

$\Longrightarrow \sf {x}^{2} {y}^{2} ( {x}^{2} + {y}^{2} + 3)$

$\Longrightarrow \sf { \bigg( \dfrac{ \cos ^{2} ( \theta) }{ \sin(\theta) } \bigg) }^{2} { \bigg( \dfrac{ \sin^{2} ( \theta) }{ \cos(\theta) } \bigg) }^{2} \bigg(\bigg( \dfrac{ \cos ^{2} ( \theta) }{ \sin(\theta) } \bigg)^{2} + \bigg( \dfrac{ \sin^{2} ( \theta) }{ \cos(\theta) } \bigg)^{2} + 3 \bigg)$

$\Longrightarrow \sf { \dfrac{ \cancel{ \cos ^{4} ( \theta) }}{ \cancel{\sin^{2} (\theta)} }} \times { \dfrac{ \cancel{\sin^{4} ( \theta)} }{ \cancel{\cos ^{2} (\theta)} }} \bigg(\dfrac{ \cos ^{4} ( \theta) }{ \sin^{2} (\theta) } + \dfrac{ \sin^{4} ( \theta) }{ \cos^{2} (\theta) } + 3 \bigg)$

$\Longrightarrow \sf \cancel{\cos ^{2} ( \theta) \sin ^{2} ( \theta) }\times \bigg(\dfrac{ \cos ^{6} ( \theta) + \sin^{6}( \theta) + 3 \sin ^{2}( \theta) \cos ^{2} ( \theta) }{ \cancel{\sin^{2}( \theta) \cos^{2} ( \theta)}} \bigg)$

$\Longrightarrow \sf \cos ^{6} ( \theta) + \sin^{6}( \theta) + 3 \sin ^{2}( \theta) \cos ^{2} ( \theta) \times (1)$

⠀⠀⠀⋆ (sin²(θ) + cos²(θ)) = 1

$\Longrightarrow \sf (\cos ^{2} ( \theta))^{3} + (\sin^{2}( \theta) ) ^{3} + 3 \sin ^{2}( \theta) \cos ^{2} ( \theta) \times (\sin^{2}( \theta) + \cos ^{2} ( \theta))$

⠀⠀⠀⋆ (a + b)³ = a³ + b³ + 3ab(a + b)

$\Longrightarrow \sf ((\cos ^{2} ( \theta)+ (\sin^{2}( \theta) ) ^{3}$

$\Longrightarrow \sf (1) ^{3}$

$\Longrightarrow \large \sf 1$

⠀

$\therefore \large \boxed{ \sf{x}^{2} {y}^{2} ( {x}^{2} + {y}^{2} + 3) = 1}$