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Question: $\sqrt{3}$ sec10°+8sin 10°cos 80° is equal to-...

3\sqrt{3} sec10°+8sin 10°cos 80° is equal to-

A

2

B

-2

C

0

D

3\sqrt{3}

Answer

2

Explanation

Solution

The given expression is 3sec⁡10∘+8sin⁡10∘cos⁡80∘\sqrt{3} \sec 10^\circ + 8 \sin 10^\circ \cos 80^\circ.

We will simplify this expression step-by-step using trigonometric identities.

  1. Rewrite sec⁡10∘\sec 10^\circ and cos⁡80∘\cos 80^\circ:
    We know that sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x} and cos⁡(90∘−x)=sin⁡x\cos(90^\circ - x) = \sin x.
    So, sec⁡10∘=1cos⁡10∘\sec 10^\circ = \frac{1}{\cos 10^\circ}.
    And cos⁡80∘=cos⁡(90∘−10∘)=sin⁡10∘\cos 80^\circ = \cos(90^\circ - 10^\circ) = \sin 10^\circ.

Substitute these into the expression: E=3cos⁡10∘+8sin⁡10∘(sin⁡10∘)E = \frac{\sqrt{3}}{\cos 10^\circ} + 8 \sin 10^\circ (\sin 10^\circ) E=3cos⁡10∘+8sin⁡210∘E = \frac{\sqrt{3}}{\cos 10^\circ} + 8 \sin^2 10^\circ

  1. Use the double angle identity for sin⁡2x\sin^2 x:
    We know that 2sin⁡2x=1−cos⁡2x2 \sin^2 x = 1 - \cos 2x.
    So, 8sin⁡210∘=4(2sin⁡210∘)=4(1−cos⁡(2×10∘))=4(1−cos⁡20∘)8 \sin^2 10^\circ = 4 (2 \sin^2 10^\circ) = 4 (1 - \cos (2 \times 10^\circ)) = 4 (1 - \cos 20^\circ).

Substitute this back into the expression for EE: E=3cos⁡10∘+4(1−cos⁡20∘)E = \frac{\sqrt{3}}{\cos 10^\circ} + 4 (1 - \cos 20^\circ) E=3cos⁡10∘+4−4cos⁡20∘E = \frac{\sqrt{3}}{\cos 10^\circ} + 4 - 4 \cos 20^\circ

  1. Combine the terms with cos⁡10∘\cos 10^\circ and cos⁡20∘\cos 20^\circ:
    To combine the first and last terms, find a common denominator: E=3−4cos⁡20∘cos⁡10∘cos⁡10∘+4E = \frac{\sqrt{3} - 4 \cos 20^\circ \cos 10^\circ}{\cos 10^\circ} + 4

  2. Use the product-to-sum identity for cos⁡Acos⁡B\cos A \cos B:
    We know that 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2 \cos A \cos B = \cos(A+B) + \cos(A-B).
    Let A=20∘A = 20^\circ and B=10∘B = 10^\circ.
    So, 4cos⁡20∘cos⁡10∘=2(2cos⁡20∘cos⁡10∘)4 \cos 20^\circ \cos 10^\circ = 2 (2 \cos 20^\circ \cos 10^\circ) =2(cos⁡(20∘+10∘)+cos⁡(20∘−10∘))= 2 (\cos(20^\circ + 10^\circ) + \cos(20^\circ - 10^\circ)) =2(cos⁡30∘+cos⁡10∘)= 2 (\cos 30^\circ + \cos 10^\circ) We know cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2}. =2(32+cos⁡10∘)= 2 \left(\frac{\sqrt{3}}{2} + \cos 10^\circ \right) =3+2cos⁡10∘= \sqrt{3} + 2 \cos 10^\circ

  3. Substitute this result back into the expression for EE: E=3−(3+2cos⁡10∘)cos⁡10∘+4E = \frac{\sqrt{3} - (\sqrt{3} + 2 \cos 10^\circ)}{\cos 10^\circ} + 4 E=3−3−2cos⁡10∘cos⁡10∘+4E = \frac{\sqrt{3} - \sqrt{3} - 2 \cos 10^\circ}{\cos 10^\circ} + 4 E=−2cos⁡10∘cos⁡10∘+4E = \frac{-2 \cos 10^\circ}{\cos 10^\circ} + 4 E=−2+4E = -2 + 4 E=2E = 2

The value of the expression is 2.