If sin -cos - p and -cosec- q- then q p 2-1 is equal toA- 3 pB- 2 pC- PD- None of these

The correct option is B 2p We have, LHS=q(p^2 1) ⇒ LHS = (sec θ + cosec θ ) { (sin θ + cos θ)^2 1 } ⇒ LHS = ( 1/cos θ+ 1/sin θ ) { sin^2 θ + cos^2 θ + 2sin θ ...

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Standard X

Mathematics

Trigonometric Identities

If sin θ+cos ...

Question

If $s i n (θ) + c o s (θ) = p a n d s e c (θ) + c o s e c (θ) = q$ , then $q (p^{2} - 1)$ is equal to

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None of these

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Solution

The correct option is B
2p

$W e h a v e, L H S = q (p^{2} - 1) \Rightarrow L H S = (s e c θ + c o s e c θ) {(s i n θ + c o s θ)^{2} - 1} \Rightarrow L H S = (\frac{1}{c o s θ} + \frac{1}{s i n θ}) {s i n^{2} θ + c o s^{2} θ + 2 s i n θ c o s θ - 1} \Rightarrow L H S = (\frac{s i n θ + c o s θ}{c o s θ s i n θ}) (1 + 2 s i n θ c o s θ - 1) \Rightarrow L H S = (\frac{s i n θ + c o s θ}{c o s θ s i n θ}) (2 s i n θ c o s θ) \Rightarrow L H S = 2 (s i n θ + c o s θ) = 2 p = R H S$

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Similar questions

Q. If (sin θ + cos θ) = p and (sec θ + cosec θ) = q, then q(p² − 1) = ?
(a) 2
(b) 2p
(c)

\frac{1}{p^{2}}

(d)

\frac{1}{p}

Q. If

sin θ + cos θ = p

and

sec θ + c o s e c θ = q

, show that

q (p^{2} - 1) = 2 p

cos θ + sin θ = p

and

sec θ + c o s e c θ = q

Prove that

q (p^{2} - 1) = 2 p

Q. Choose the correct answer from the alternatives given.
If

sin θ, + cos θ = p

and

sec θ, + cosec θ = q

. then

q (p^{2}

- 1)

= ?

If $c o s e c θ - s i n θ = a^{3}$ and $s e c θ - c o s θ = b^{3}$ , then $a^{2} b^{2} (a^{2} + b^{2}) =$

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If sin(θ) + cos(θ)=p and sec(θ)+cosec(θ)= q, then q(p2–1) is equal to

The correct option is B 2p We have,LHS=q(p2−1)⇒LHS=(secθ+cosecθ){(sinθ+cosθ)2−1}⇒LHS=(1cosθ+1sinθ){sin2θ+cos2θ+2sinθcosθ−1}⇒LHS=(sinθ+cosθcosθsinθ)(1+2sinθcosθ−1)⇒LHS=(sinθ+cosθcosθsinθ)(2sinθcosθ)⇒ LHS=2(sinθ+cosθ)=2p=RHS

If $s i n (θ) + c o s (θ) = p a n d s e c (θ) + c o s e c (θ) = q$ , then $q (p^{2} - 1)$ is equal to