I f y = √ ( 1 + c o s 2 θ 1 − c o s 2 θ ) , d y d θ a t θ = 3 π 4 i s Options: A . &nbsp − 2 B . &nbsp 2 C . &nbsp ± 2 D . &nbsp None of these

Answer: Option B : B y = √ ( 1 + c o s 2 θ 1 − c o s 2 θ ) = | c o t θ | = − c o t θ ( θ = 3 π 4 ) ∴ d y d θ = c o s e c 2 θ d y d θ | θ = 3 π / 4 = ( √ 2 ) 2 = 2

If y=√(1+cos 2 θ1−cos2 θ),dyd&#x

Question

$I f y = \sqrt{(\frac{1 + c o s 2 θ}{1 - c o s 2 θ})}, \frac{d y}{d θ} a t θ = \frac{3 π}{4} i s$

Options:

A .

- 2

B .

2

C .

\pm 2

D . None of these

Answer: Option B
:
B

y = \sqrt{(\frac{1 + c o s 2 θ}{1 - c o s 2 θ})}

= | c o t θ | = - c o t θ (θ = \frac{3 π}{4})

∴ \frac{d y}{d θ} = c o s e c^{2} θ

\frac{d y}{d θ} |_{θ = 3 π / 4} = (\sqrt{2})^{2} = 2

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