If x - acosec2- - sec- and y - bsec- - cot2- where 0- - - - 90- then the correct option is- x - acosec2- - sec- - y - bsec- - cot2- - 0- - - - 90- - - - -

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Parametric Differentiation

If x = acosec...

Question

If x = a(cosec²θ – secθ) and y = b(secθ – cot²θ) (where 0° < θ < 90°), then the correct option is
यदि x = a(cosec²θ – secθ) तथा y = b(secθ – cot²θ) (जहाँ 0° < θ < 90°), तब सही विकल्प है

\frac{x}{a} + \frac{y}{b} = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

xy + ab = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x²y² = ab

No worries! We‘ve got your back. Try BYJU‘S free classes today!

xy – ab = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{x}{a} + \frac{y}{b} = 1$
Solution

Suggest Corrections

Similar questions

Q. If tan

θ

and cot

θ

are the roots of the equation x² – ax + a² – 4 = 0 (0° <

θ

< 90°), then the correct option(s) is/are

यदि समीकरण x² – ax + a² – 4 = 0 (0° <

θ

< 90°) के मूल tan

θ

तथा cot

θ

हैं, तब सही विकल्प है/हैं

c o t^{2} - c o s e c^{2} θ =

...............

(0 < θ < 90)

Q. This section contains 1 Assertion-Reason type question, which has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

इस खण्ड में 1 कथन-कारण प्रकार का प्रश्न है, जिसमें 4 विकल्प (a), (b), (c) तथा (d) दिये गये हैं, जिनमें से केवल एक सही है।

A : If cosecθ – cotθ = q (where q ≠ 0), then the value of cosec³θ – cot³θ is

\frac{3 - q^{4}}{4 q}

, where 0° < θ < 90°.

A : यदि cosecθ – cotθ = q (जहाँ q ≠ 0), तब cosec³θ – cot³θ का मान

\frac{3 - q^{4}}{4 q}

है, जहाँ 0° < θ < 90° है।

R : cosec²θ – cot²θ = 1.

Q. If tan

θ

+ cot

θ

= x, where x is the least prime number and tan

ϕ

+ sec

ϕ

= a, where 0° <

ϕ

θ

< 90°, then the correct statement(s) is/are
Statement (I) : The value of

sinϕ = \frac{a^{2} - 1}{a^{2} + 1}

.
Statement (II) : The value of

cosϕ (a + \frac{1}{a})

is a prime number.
Statement (III) : The value of

\frac{1}{\sqrt{3}} (sinθ + cosθ) = x

.

यदि tan

θ

+ cot

θ

= x, जहाँ x लघुत्तम अभाज्य संख्या है तथा tan

ϕ

+ sec

ϕ

= a, जहाँ 0° <

ϕ

θ

< 90°, तब सही कथन है/हैं
कथन (I) : sin

ϕ

का मान =

sinϕ = \frac{a^{2} - 1}{a^{2} + 1}

कथन (II) :

cosϕ (a + \frac{1}{a})

का मान एक अभाज्य संख्या है।
कथन (III) :

\frac{1}{\sqrt{3}} (sinθ + cosθ)

का मान = x

Q. If x = θcosθ – sinθ and y = θsinθ + cosθ, where 'θ' is a parameter, then

\frac{d y}{d x}

equals

यदि x = θcosθ – sinθ तथा y = θsinθ + cosθ, जहाँ 'θ' एक प्राचल है, तब

\frac{d y}{d x}

का मान है

Join BYJU'S Learning Program

If x = a(cosec2θ – secθ) and y = b(secθ – cot2θ) (where 0° < θ < 90°), then the correct option is यदि x = a(cosec2θ – secθ) तथा y = b(secθ – cot2θ) (जहाँ 0° < θ < 90°), तब सही विकल्प है

The correct option is A xa+yb=1Solution

If x = a(cosec²θ – secθ) and y = b(secθ – cot²θ) (where 0° < θ < 90°), then the correct option is
यदि x = a(cosec²θ – secθ) तथा y = b(secθ – cot²θ) (जहाँ 0° < θ < 90°), तब सही विकल्प है

The correct option is A $\frac{x}{a} + \frac{y}{b} = 1$
Solution