Today i am solving introduction to trigonometry question of exercise 8.1 of question no. 6 but i can’t solve this question .Find the best way to solve this question in a easy way its so important question for this chapter.If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

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Let us assume the triangle ABC in which CD⊥AB

Give that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cos ratio is written as

AD/AC = BD/BC

Now, interchange the terms, we get

AD/BD = AC/BC

Let take a constant value

AD/BD = AC/BC = k

Now consider the equation as

AD = k BD …(1)

AC = k BC …(2)

By applying Pythagoras theorem in △CAD and △CBD we get,

CD

^{2}= BC^{2}– BD^{2 }… (3)CD

^{2 }=AC^{2 }−AD^{2}….(4)From the equations (3) and (4) we get,

AC

^{2}−AD^{2 }= BC^{2}−BD^{2}Now substitute the equations (1) and (2) in (3) and (4)

K

^{2}(BC^{2}−BD^{2})=(BC^{2}−BD^{2}) k^{2}=1Putting this value in equation, we obtain

AC = BC

∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)