Mathematics Multiple Choice Questions on “Properties of Inverse Trigonometric Functions”.
1. sin-1x in terms of cos-1is ____________
a) cos-1(sqrt{1+x^2})
b) cos-1(sqrt{1-x^2})
c) cos-1x
d) cos-1(frac{1}{x})
Answer: b
Clarification: Let sin-1x=y
⇒x=siny
⇒(x=sqrt{1-cos^2y})
⇒(x^2=1-cos^2y)
⇒(cos^2y=1-x^2)
∴y=cos-1(sqrt{1-x^2})=sin-1x.
2. What is sec-1x in terms of tan-1?
a) tan-1(sqrt{1+x^2})
b) tan-11+x2
c) tan-1x
d) tan-1(sqrt{x^2-1})
Answer: d
Clarification: Let sec-1x=y
⇒x=secy
⇒x=(sqrt{1+tan^2y})
⇒x2-1=tan2y
∴y=tan-1(sqrt{x^2-1})=sec-1x.
3. What is the value of cos(tan-1((frac{4}{5})))?
a) (frac{5}{4})
b) (frac{5}{sqrt{41}})
c) (frac{sqrt{41}}{5})
d) (frac{4}{5})
Answer: b
Clarification: From ∆ABC, we get
tan-1((frac{4}{5}))=cos-1((frac{5}{sqrt{41}}))
cos(tan-1((frac{4}{5}))=cos(cos-1((frac{5}{sqrt{41}})))
=(frac{5}{sqrt{41}})
4. What is the solution of cot(sin-1x)?
a) (frac{sqrt{1-x^2}}{x})
b) x
c) (sqrt{1-x^2})
d) (sqrt{1+x^2})
Answer: a
Clarification: Let sin-1x=y. From ∆ABC, we get
y=sin-1x=cot-1((frac{sqrt{1-x^2}}{x}))
∴cot(sin-1x)=(cot(cot^{-1}(frac{sqrt{1-x^2}}{x}))=frac{sqrt{1-x^2}}{x}).
5. Which of the following formula is incorrect?
a) sin-1x+sin-1y=sin-1{(xsqrt{1-y^2}+ysqrt{1-x^2})}
b) sin-1x-sin-1y=sin-1{(xsqrt{1+y^2}+ysqrt{1+x^2})}
c) 2 tan-1x=tan-1((frac{2x}{1-x^2}))
d) 2 cos-1x=cos-1(3x-4x3)
Answer: b
Clarification: The formula sin-1x-sin-1y=sin-1{(xsqrt{1+y^2}+ysqrt{1+x^2})} is incorrect. The correct formula is sin-1x-sin-1y=sin-1{(xsqrt{1+y^2}-y sqrt{1-x^2})}.
6. Find the value of sin-1((frac{5}{13}))+cos-1((frac{3}{5})).
a) sin-1((frac{63}{65}))
b) sin-11
c) 0
d) sin-1((frac{64}{65}))
Answer: a
Clarification: From ∆ABC, we get
cos-1((frac{3}{5}))=sin-1((frac{4}{5}))
∴sin-1((frac{5}{13}))+cos-1((frac{3}{5}))=sin-1((frac{5}{13}))+sin-1((frac{4}{5}))
=sin-1((frac{5}{13}sqrt{1-(frac{4}{5})^2}+frac{4}{5}sqrt{1-(frac{5}{13})^2}))
=(sin^{-1}(frac{5}{13}×frac{3}{5}+frac{4}{5}×frac{12}{13})=sin^{-1}(frac{15+48}{65})=sin^{-1}(frac{63}{65})).
7. Find the value of tan-1((frac{1}{3}))+tan-1((frac{1}{5}))+tan-1(frac{1}{7})[/latex]
a) tan-1((frac{4}{7}))
b) tan-1((frac{9}{7}))
c) tan-1((frac{7}{9}))
d) tan-11
Answer: c
Clarification: Using the formula tan-1x+tan-1y=tan-1(frac{x+y}{1-xy}), we get
tan-1((frac{1}{3}))+tan-1((frac{1}{5}))=tan-1(bigg(frac{frac{1}{3}+frac{1}{5}}{1-frac{1}{3}×frac{1}{5}}bigg))
= (tan^{-1}bigg(frac{frac{8}{15}}{frac{14}{15}}bigg)=tan^{-1}(frac{8}{15}×frac{15}{14})=tan^{-1}(frac{4}{7}))
=(tan^{-1}(frac{1}{3})+tan^{-1}(frac{1}{5})+tan^{-1}(frac{1}{7})=tan^{-1}(frac{4}{7})+tan^{-1}(frac{1}{7}))
=(tan^{-1}bigg(frac{frac{4}{7} + frac{1}{7}}{1-frac{4}{7}×frac{1}{7}}bigg) = tan^{-1}bigg(frac{frac{5}{7}}{frac{45}{49}}bigg)=tan^{-1}(frac{5}{7}×frac{49}{45}))
=tan-1((frac{7}{9})).
8. Find the value of sin-1((frac{3}{5}))+sin-1((frac{4}{5}))+cos-1((frac{sqrt{3}}{2})).
a) (frac{π}{3})
b) (frac{2π}{3})
c) (frac{4π}{3})
d) (frac{π}{4})
Answer: b
Clarification: Using the formula sin-1x+sin-1y=sin-1({x sqrt{1-y^2}+y sqrt{1-x^2}}), we get
sin-1((frac{3}{5}))+sin-1((frac{4}{5}))=sin-1(Big{ frac{3}{5} sqrt{1-(frac{4}{5})^2}+frac{4}{5} sqrt{1-(frac{3}{5})^2} Big})
=sin-1((frac{3}{5})×(frac{3}{5})+(frac{4}{5})×(frac{4}{5}))=sin-1((frac{25}{25})=frac{π}{2})
∴ sin-1((frac{3}{5}))+sin-1((frac{4}{5}))+cos-1((frac{sqrt{3}}{2})=frac{π}{2}+frac{π}{6}=frac{3π+π}{6}=frac{2π}{3}).
9. What is the value of 2 tan-1x in terms of sin-1?
a) sec-1x
b) 2 sec-1x
c) 2 sec-1((sqrt{1+x^2}))
d) sec-1((sqrt{1+x^2}))
Answer: c
Clarification: Let 2 tan-1x=y
⇒tan-1x=(frac{y}{2})
From ∆ABC, we get
⇒tan-1x=sec-1(sqrt{1+x^2}=frac{y}{2})
⇒y=2 sec-1((sqrt{1+x^2}))
10. sin-1x+cos1x= ___
a) (frac{π}{2})
b) π
c) (frac{π}{3})
d) 2π
Answer: a
Clarification: sin-1x+cos-1x=(frac{π}{2}); x∈[-1,1]