Finding the value of sin−11213+cos−145+tan−16313

Alan Smith

Alan Smith

Answered

2022-01-17

Finding the value of sin11213+cos145+tan16313

Answer & Explanation

RizerMix

RizerMix

Expert

2022-01-19Added 437 answers

Actually
tan1125+tan134==π+tan1(125+34112534)
We can notice that
π2<tan1125+tan134<π
I am going to prove that
NSK
if π2<α+β<3π2 then α+β=π+tan1(tanα+tanβ1tanαtanβ)
Proof:
tan(α+βπ)=tanα+tan(βπ)1tanαtan(βπ)=tanα+tanβ1tanαtanβ
As |α+βπ|<π2 and the function tangent is invertible in [π2,π2] it follows that
α+βπ=tan1(tanα+tanβ1tanαtanβ) , therefore:
α+β=π+tan1(tanα+tanβ1tanαtanβ)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?