Which of the following options are equal -1-cos A -1-cos A -A- 1-cos A -sin AB- 1-cos A -sin AC- cosec A- AD- A-tan A

The correct options are B 1 + cos A/sin A C cosec A + cot A √(1 + cos A/1 cos A) = √((1+cos A/1 cos A) (1+cos A/1+cos A)) = √((1 + cos A)^2/1 cos^2 A) = √((1 ...

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Standard X

Mathematics

Trigonometric Identities

Which of the ...

Question

Which of the following options are equal to $\sqrt{\frac{1 + c o s A}{1 - c o s A}} ?$

$\frac{1 - c o s A}{s i n A}$

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$\frac{1 + c o s A}{s i n A}$

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$c o s e c A + c o t A$

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$s e c A + t a n A$

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Solution

The correct options are
B
$\frac{1 + c o s A}{s i n A}$

C
$c o s e c A + c o t A$

$\sqrt{\frac{1 + c o s A}{1 - c o s A}} = \sqrt{(\frac{1 + c o s A}{1 - c o s A}) (\frac{1 + c o s A}{1 + c o s A})}$

$= \sqrt{\frac{(1 + c o s A)^{2}}{1 - c o s^{2} A}}$

$= \sqrt{\frac{(1 + c o s A)^{2}}{s i n^{2} A}} . . . . (s i n^{2} A + c o s^{2} A = 1)$

$= \frac{1 + c o s A}{s i n A}$

$= \frac{1}{s i n A} + \frac{c o s A}{s i n A}$

$= c o s e c A + c o t A$

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Similar questions

\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{cosec A - 1}{cosec A + 1}

Q. Prove that

\frac{cot A - cos A}{cot A + cos A} = \frac{cos e c A - 1}{cos e c A + 1}

Q. Prove the following trigonometric identities.

\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{cosec A - 1}{cosec A + 1}

(cosec A - sin A) (sec A - cos A) = \frac{1}{tan A + cot A}

Prove the following identities:

$\frac{s i n A}{s e c A + t a n A - 1} + \frac{c o s A}{c o s e c A + c o t A - 1} = 1$

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Which of the following options are equal to √1+cos A1−cos A ?

The correct options are B 1+cos Asin A C cosec A+cot A √1+cos A1−cos A=√(1+cos A1−cos A)(1+cos A1+cos A) =√(1+cos A)21−cos2A =√(1+cos A)2sin2 A....(sin2A+cos2A=1) =1+cos AsinA =1sin A+cos Asin A =cosec A+cot A

Which of the following options are equal to $\sqrt{\frac{1 + c o s A}{1 - c o s A}} ?$