If 3 π 4 < α < π , then √ c o s e c 2 α + 2 c o t α is equal to Options: A . &nbsp 1 + c o t α B . &nbsp 1 − c o t α C . &nbsp − 1 − c o t α D . &nbsp − 1 + c o t α

Answer: Option C : C √ c o s e c 2 α + 2 c o t α = √ 1 + c o t 2 α + 2 c o t α = | 1 + c o t α | But 3 π 4 < α < π ⇒ c o t α < − 1 ⇒ 1 + c o t α < 0 Hence, | 1 + c o t α | = − ( 1 + c o t α )

If 3π4<α<π, then √cosec2α+2cotα is

Discussion Forum : Trigonometric Functions

Question -

If $\frac{3 π}{4} < α < π$ , then $\sqrt{c o s e c^{2} α + 2 c o t α}$ is equal to

Options:

A .

1 + c o t α

B .

1 - c o t α

C .

- 1 - c o t α

D .

- 1 + c o t α

Answer: Option C
:
C

$\sqrt{c o s e c^{2} α + 2 c o t α} = \sqrt{1 + c o t^{2} α + 2 c o t α} = | 1 + c o t α |$
But $\frac{3 π}{4} < α < π \Rightarrow c o t α < - 1 \Rightarrow 1 + c o t α < 0$
Hence, $| 1 + c o t α | = - (1 + c o t α)$

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Discussion Forum : Trigonometric Functions

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