Math, asked by anirudh53187, 11 months ago

the value of (tan1,tan2,tan3,.....tan89is one proof brainliest can solve​

Answers

Answered by Anonymous
15

\Huge{\underline{\underline{\mathfrak{Answer \colon }}}}

Given,

tan1°,tan2°...................................tan89°

This's not gcan be rewritten as :

tan1.tan2.tan3...............tan45................tan89

Since,

 \sf{tan (90 - \theta) \:  =cot  \theta }

For instance,

 \sf{tan(90 - 89) = cot \: 1}

Similarly,we can write the given sequence as :

tan1.tan2.tan3.........tan45................cot3.cot2.cot1

Also,

 \sf{tan \: x =  \frac{1}{cot \: x} } \\

Thus,all the terms cancel out eachother leaving,

tan45

= 1

Thus,

tan1.tan2.tan3.................tan89 = 1

Answered by Nereida
3

\huge\green{\green{\underline{\mathfrak{Question :-}}}}

Prove that tan 1,tan 2,tan 3.....tan 89 = 1.

\huge\green{\green{\underline{\mathfrak{Answer :-}}}}

tan 1 \times tan 2 \times........tan 45.....tan 89

Formula :- tan(90 - theta) = cot( theta).

So,

Let us write tan 89, tan 88.... as ...

tan (90 - 89) = cot 1

tan (90 - 88) = cot 2....and so on till tan 46.

So, it becomes...

tan 1 \times tan 2 \times......tan 45.....\\...cot 3 \times cot 2 \times cot 1.

We also know that :- cot theta is the reciprocal of tan theta.

So,

We can rewrite it as :-

\cancel {tan 1} \times \cancel {tan 2} \times \cancel {tan 3}.......tan 45.....\\...\dfrac {1}{\cancel {tan 3}} \times \dfrac {1}{\cancel {tan 2}} \times \dfrac{1}{\cancel {tan 1}}.

So, only tan 45 will be left.

We know that, tan 45 = 1.

Therefore, our answer is 1.

Hence, proved.

____________________

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