Math, asked by Grsoul5341, 1 year ago

prove that 1/sin1sin2 + 1/sin2sin3 +......+1/sin89sin90 = cos1/sin^21

Answers

Answered by CarlynBronk
6

Solution:

LHS

\frac{1}{sin 1 sin 2}+\frac{1}{sin 2 sin 3}+\frac{1}{sin 3 sin 4}+\frac{1}{sin 4 sin 5}+........\frac{1}{sin 89 sin 90}

Now,

\frac{1}{sin 1 sin 2}=\frac{sin (2-1)}{sin 1 *sin 2}=\frac{sin 2 *cos 1 - sin 1* cos 2}{sin 1 * sin 2}=cot 1 - cot 2

Using the trigonometric identity

sin (a+b)= sin a cos b +cos a sin b

sin(a-b)=sin a cos b - cos a sin b

Similarly, other terms can be written as

cot 2-cot 3, cot 3-cot 4,........... , cot 89-cot 90.

So, adding all these terms , we get

→→cot 1 - cot 2 +cot 2-cot 3 +cot 3 - cot 4 +cot 4 - cot 5 +........+cot 88 - cot 89+cot 89 -cot 90

Cancelling out the like terms

→cot 1 - cot 90

=\frac{cos 1}{sin 1}-\frac{cos 90}{sin 90}\\\\=\frac{cos 1*sin 90 -sin1* cos 90}{sin 1 * sin 90}\\\\=\frac{sin (90-1)}{sin 1*sin 90}\\\\=\frac{sin 89}{sin 1 * sin 90}\\\\ =\frac{cos 1}{sin 1 * sin 90}{\text{as, sin(90-a)=cos a}}, so [sin (90-1)]=cos 1]

As, sin 1 and sin 90 Radian, are approximately same, so you can write it as

=\frac{cos 1}{sin 1 *sin 1}\\\\==\frac{cos 1}{sin^2 1 }

= R H S

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