If 3 -4- then -2 -1-sin 2- is equal to

The correct option is B √(2 cot α1/sin^2 α)=√(2 cot α+cos ec^2 α)=√(1+cot^2α + 2 cot α) |1+cot α|since cot α lt; 1 when 3 π/4 lt; α lt; π, we have | ...

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Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

If 3 π/4<α<π,...

Question

$I f \frac{3 π}{4}$ < $α$ < $π$ , then $\sqrt{2 c o t α \frac{1}{s i n^{2} α}}$ is equal to

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Solution

The correct option is B

$\sqrt{2 c o t α \frac{1}{s i n^{2} α}} = \sqrt{2 c o t α + c o s e c^{2} α} = \sqrt{1 + c o t^{2} α + 2 c o t α}$

$| 1 + c o t α | s i n c e c o t α$ < -1 when $\frac{3 π}{4}$ < $α$ < $π$ , we have $| 1 + c o t α | = - 1 - c o t α$

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Similar questions

If $c o s A = \frac{4}{5}$ and $c o s B = \frac{12}{13}, \frac{3 π}{2} < A, B < 2 π$ , find the value of the following

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Q. If

f (x) = ⎧ ⎨ ⎩ \begin{matrix} 1, w h e n 0 < x \leq \frac{3 π}{4} 2 s i n \frac{2}{9} x, w h e n \frac{3 π}{4} < x < π \end{matrix}

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A pole of height 12 feet is reduced with a scale

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\frac{3 + 2 i s i n θ}{1 - 2 i s i n θ}

will be purely imaginary, if

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Q. For any angle

A, \sqrt{1 + s i n A} = ∣ ∣ c o s \frac{A}{2} + s i n \frac{A}{2} ∣ ∣, \sqrt{1 - s i n A} = ∣ ∣ c o s \frac{A}{2} - s i n \frac{A}{2} ∣ ∣

Use this information to answer the following.

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If 3π4 < α < π, then √2 cot α1sin2α is equal to

The correct option is B √2 cot α1sin2α=√2 cot α+cos ec2α=√1+cot2α+2cotα |1+cotα|since cot α < -1 when 3π4 < α < π, we have |1+cot α|=−1−cot α

$I f \frac{3 π}{4}$ < $α$ < $π$ , then $\sqrt{2 c o t α \frac{1}{s i n^{2} α}}$ is equal to

The correct option is B

$\sqrt{2 c o t α \frac{1}{s i n^{2} α}} = \sqrt{2 c o t α + c o s e c^{2} α} = \sqrt{1 + c o t^{2} α + 2 c o t α}$

$| 1 + c o t α | s i n c e c o t α$ < -1 when $\frac{3 π}{4}$ < $α$ < $π$ , we have $| 1 + c o t α | = - 1 - c o t α$