Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sin inverse X + Cot inverse X + tan inverse X Range=????
we know the standard identity:Cot inverse x + tan inverse x= pi/2also, Sin inverse x lies in [– pi/2, pi/2].so that range of Sin inverse X + Cot inverse X + tan inverse X= Sin inverse X + pi/2 is [0, pi].KINDLY APPROVE :))
f(x)=sin−1x+cos−1x+tan−1xUsing the Inverse Trigonometric Identity,sin−1x+cos−1x=π/2We get,f(x)=π/2+tan−1xNow,, range of tan−1x as [−π/2,π/2]…-π/2 so,-π/2+π/2 here it comes , 0 but here it comes a great mistake becauseNow most of us apply the range of tan−1x as [−π/2,π/2]So we get the range of f(x) as [0,π] but there is a very big mistake here.We know that the domain of f(x)f(x) is [−1,1] hence the term tan−1x wouldn’t hold the values −π/2 and π/2So the new range becomes,f(−1)=π/2−π/4=π/4 f(−1)=π2−π/4=π/4f(1)=π2+π/4=3π/4 f(1)=π/2+π/4=3π/4So we get f(x)∈[π/4,3π/4]
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -