If A -280- then -1-sin A -1-sin A -A- 2 sin 50 oB- 2 cos 50-C- 2 sin 100D- 2 cos 10-

The correct option is A 2 sin50 If A = 280^∘, then √((1+sinA)) + √((1 sinA)) = | cos A/2 + sin A/2 | + | cos A/2 sin A/2 | = | cos 140^∘ + sin 140^∘ | + | ...

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Standard XII

Mathematics

Differentiation of Inverse Trigonometric Functions

If A =280∘ th...

Question

If $A = 280^{\circ}$ then $\sqrt{(1 + s i n A)}$ + $\sqrt{(1 - s i n A)}$ =

2 sin50

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2 cos50

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2 sin10

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2 cos10

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Solution

The correct option is A
2 sin50

If $A = 280^{\circ}$ , then

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

= $| c o s 140^{\circ} + s i n 140^{\circ} |$ + $| c o s 140^{\circ} - s i n 140^{\circ} |$

= $| - s i n 50^{\circ} + c o s 50^{\circ} |$ + $| - s i n 50^{\circ} - c o s 50^{\circ} |$

= $s i n 50^{\circ} - c o s 50^{\circ} + s i n 50^{\circ} + c o s 50^{\circ}$ = $2 s i n 50^{\circ}$

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Similar questions

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.
If

A = 280^{\circ}

then

\sqrt{1 + s i n A} + \sqrt{1 - s i n A} =

If A = 340^{o}, then

\sqrt{1 - sin A} - \sqrt{1 + sin A}

is equal to

Q. If

s i n (B + C) = \frac{\sqrt{3}}{2}

s i n (A + C) = s i n B and

s i n A = \frac{1}{2} .

, then

sin (A + B - C)

is .

Q. If

\sqrt{\frac{1 - sin A}{1 + sin A}} + \frac{sin A}{cos A} = \frac{1}{cos A}

, for all permissible values of

A

, then

A

belongs to

Q. If

\sqrt{\frac{1 - sin A}{1 + sin A}} = sec A - tan A,

then

A

lies in the quadrants

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If A=280∘ then √(1+sinA) + √(1−sinA) =

The correct option is A 2 sin50 If A=280∘, then √(1+sinA) + √(1−sinA) = ∣∣cosA2+sinA2∣∣ + ∣∣cosA2−sinA2∣∣ = |cos140∘+sin140∘| + |cos140∘−sin140∘| = |−sin50∘+cos50∘| + |−sin50∘−cos50∘| = sin50∘ − cos50∘ + sin50∘ + cos50∘ = 2sin50∘

If $A = 280^{\circ}$ then $\sqrt{(1 + s i n A)}$ + $\sqrt{(1 - s i n A)}$ =