Math Chapter 2 Inverse Trigonometric Functions
Chapter 2: Inverse Trigonometric Functions Exercise 2.1 Q1. Find principal value for
sin−1(−12)
Soln:
Let sin−1(−12) = a, then
– π6) sina=−12= – sin sinπ6=sin( – We know, The principal value branch range for sin-1 is [−π2,π2] and sin(−π6)= – 12 12 Therefore principal value for sin−1(−12)is – π6
Q2. Find principal value for
– 3√2) cos−1( –
Soln:
Let cos−1( – – 3√2) = a, then
cosa=3√2=cos(π6) We know, The principal value branch range for cos-1 is [0,π] and cos(π6)=3√2 Therefore, principal value for cos−1( – – 3√2)isπ6
Q3. Find principal value for cosec -1 (2) Soln:
Let cosec-1 (2) = a. Then, cosec a = 2 = cosec (π6) We know, The principal value branch range for cosec-1 is [−π2,π2] – – 0 and cosec(π6) = 2 Therefore, principal value for cosec-1 (2) is π6
Q4. Find principal value for Soln:
– 3 –√) tan−1( –
Let tan−1( – – 3 –√)=a Then, tan= – 3 –√= – tan tanπ3tan(−π3) We know, The principal value branch range for tan−1is[−π2,π2]andtan(−π3)=−3 –√ Therefore, principal value for tan−1( – – 3 –√)is−π3
Q5. Find principal value for
cos−1(−12)
Soln:
Let cos−1(−12) = a, Then cosa=−12=−cosπ3=cos(π– π3)=cos( 2π3) We know, The principal value branch range for cos−1is[0,π]andcos(2π3)= – 12 12 Therefore, principal value for cos−1(−12)is2π3
Q6. Find principal value for
tan−1(−1)
Soln:
Let tan−1(−1)=a, Then, tan a = -1 = −tan(π4)=tan( – – π4) We know, The principal value branch range for tan−1is(−π2,π2)andtan( – – π4)=−1 Therefore, principal value for tan−1(−1)is−π4
Q7. Find principal value for
sec−1(23√)
Soln:
Let sec−1(23√)=a, Then seca=23√=sec(π6) We know, The principal value branch range for sec−1is[0,π] – – {π2}andsec( π6)=23√ Therefore, principal value for sec−1(23√)is;π6
Q8. Find principal value for
cot−13 –√
Soln:
Let cot−13 –√=a, Then cota=3 –√=cot(π6) We know, The principal value branch range for cot -1 is (0,π) and cot(π6)=3 –√
Therefore, principal value for cot−13 –√=π6
Q9. Find principal value for
12√) – 12 cos−1( –
Soln: 12√)=a Let cos−1( – – 12
Then cosa=−12√= – cos( cos(π4)=cos(π– π4)=cos(3π4) We know, The principal value branch range for cos-1 is [0,π]andcos(3π4)=−12√ Therefore, principal value for cos−1( – – 12 12√)is3π4 Q10. Find principal value for cosec -1 (−2 –√) Soln:
Let cosec-1(−2 –√) = a, Then cosec a = −2 –√ = -cosec(π4) = cosec (−π4) We know, The principal value branch range for cosec-1 is [−π2,π2] – – {0} {0} and cosec−π4=−2 –√ Therefore, principal value for cosec-1 (−2 –√)is−π4 Q11. Solve tan−1(1)+cos −1(−12)+sin−1(−12) Soln:
Let tan−1(1)=a, then
tana=1=tanπ4
We know, The principal value branch range for tan−1is(−π2,π2)
tan−1(1)=π4 Let cos−1(−12)=b, then
cosb=−12=−cosπ3=cos(π– π3)=cos(2π3) We know, The principal value branch range for cos-1 is [0,π]
cos−1(−12)=2π3 Let sin−1(−12)=c, then 12= – sin sinc= – 12 sinπ6=sin(−π6)
We know, The principal value branch range for sin−1is[−π2,π2]
sin−1(−12)= – π6 Now
tan−1(1)+cos−1(−12)+sin−1(−12) =π4+2π3 – π6=3π+8π– 2π12=9π12=3π4
Q12. Solve cos−1(12)+2sin −1(12) Soln:
Let cos−1(12)=a, then cosa=12=cosπ3 We know, The principal value branch range for cos-1 is [0,π]
cos−1(12)=π3 Let sin−1(−12)=b, then
sinb=12=sinπ6
We know, The principal value branch range for sin−1is[−π2,π2]
sin−1(12)=π6 Now,
cos−1(12)+2sin−1(12) =π3+2×π6=π3+π3=2π3
Q13. If sin-1 a = b, then (i) 0≤ b≤π (ii) −π2≤ b≤π2 (iii) 0
Given sin-1 a = b We know, The principal value branch range for sin−1is[−π2,π2] Therefore, −π2≤ b≤π2 Q14. The value of tan−13 –√– sec sec−1(−2) is (i) π (ii) – π3 (iii) π3 (iv) 2π3 Soln:
Let tan−13 –√=a, then
tana=3 –√=tanπ3 We know The principal value branch range for tan−1is(−π2,π2)
tan−13 –√=π3 Let sec-1(-2) = b, then sec b = -2 = – sec secπ3=sec(π– π3)=sec(2π3) We know
– {π2} The principal value branch range for sec-1 is [0,π] – sec−1(−2)=2π3
Now,
tan−13 –√– sec sec−1(−2)=π3 – 2π3= – π3 Hence option (ii) is correct
Exercise 2.2
Q1. Show that 3sin−1=sin−1(3x – 4x 4x3),x∈[−12,12] Soln:
To show: 3sin−1=sin−1(3x – 4x 4x3),x∈[−12,12] Let sin-1x = Ɵ, then x = sin Ɵ We get, RHS = sin−1(3x – 4x 4x3)=sin−1(3sinΘ– 4sin 4sin3Θ) = sin−1(sin3Θ)=3Θ=3sin−1x = LHS
3x),x∈[12,1] Q2. Show that 3cos−1x=cos−1(4x3 – 3x),x Soln:
To show: 3cos−1x=cos−1(4x3 – 3x),x 3x),x∈[12,1] Let cos-1 x = Ɵ, then x = cos Ɵ We get, RHS = cos−1(4x3 – 3x)=cos 3x)=cos−1(4cos3Θ– 3cos 3cosΘ) = cos−1(cos3Θ)=3Θ=3cos−1x = LHS
Q3. Show that tan−1211+tan−1724=tan−112
Soln:
To show: tan−1211+tan−1724=tan−112 LHS = tan−1211+tan−1724
211×724)=tan−1(48+7711×2411×24 – 1411×24 1411×24) =tan−1(211+7241 – 211
14=tan−1125251=tan−112 = RHS =tan−148+77264 – 14
Q4. Show that 2tan−112+tan−117=tan−13117 Soln:
To show: 2tan−112+tan−117=tan−13117 LHS = 2tan−112+tan−117
=tan−1[2×121 – (12)2]+tan−117=tan−11(34)+tan−117 =tan−143+tan−117=tan−1(43+171 – 43×17) =tan−1(28+33×73×7 −43×7)=tan −128+321 – 4=tan−13117=RHS
Q5. Find simplest form for
1a,a≠0 tan−11+a2√– 1a
Soln: 1a Given tan−11+a2√– 1a
Let a = tan Ɵ =tan−11+a2√– 1a 1a = tan−11+tan2Θ√– 1tanΘ =tan−1(secΘ– 1tanΘ)=tan−1(1 – cosΘsinΘ) tan−1(2sin2Θ22sinΘ2cosΘ2)=tan−1(tanΘ2) =Θ2=12tan−1a
Q6. Find the simplest form for
tan−11a2−1√, |a|> 1
Soln:
Given tan−11a2−1√
Let a = csc Ɵ tan−11a2−1√=tan−11csc2Θ−1√ =tan−11cotΘ=tan−1tanΘ=Θ=csc−1a =π2 – sec sec−1a
Q7. Find simplest form for Soln:
cosa1+cosa−−−−−√),a<π, tan−1(1 – cosa1+cosa
cosa1+cosa−−−−−√ ) Given tan−1(1 – cosa1+cosa
Now,
tan−1(1 –
−−−−−√)=tan−1(2sin2x22cos2x2−−−−−−√) tan−1(tan2x2−−−−−√)=tan−1(tanx2)
cosa1+cosa
=x2
Q8. Find simplest form for
sinacosa+sina),0
Soln:
Given tan−1(cosa – sinacosa+sina sinacosa+sina) Now, sinacosa1+sinacosa)=tan−1(1 – tana1+tana tan−1(cosa – sinacosa+sina sinacosa+sina)=tan−1(1 – sinacosa tana1+tana )
= tan−1(1 – tana1+1.tana tana1+1.tana)=tan−1(tanπ4 – tana1+tan tana1+tanπ4.tana) = tan−1[tan(π4 – a)]= a)]=π4 – a
Q9: Find simplest form for
tan−1ax2 – a2√,|a|
Soln:
Given: tan−1ax2 – a2√
Let a = x sin Ɵ tan−1ax2 – a2√=tan−1(xsinΘx2 – x2sin2Θ√)=tan−1(xsinΘx1 – sin sin2Θ√) = tan−1(xsinΘxsinΘ)=tan−1(tanΘ)=Θ=sin−1ax
3xa2),x>0;−x3√≤ax3√ Q10. Find simplest form for tan−1(3x2a – a3x3 – 3xa
Soln:
3xa2) Given tan−1(3x2a – a3x3 – 3xa
Let a = x tan Ɵ 3xa2)=tan −1(3x2.xtanΘ– x3tan3Θx3 – 3x.x 3x.x2tan2Θ) tan−1(3x2a – a3x3 – 3xa 3x3tan2Θ)=tan −1(3tanΘ– tan tan3Θ1 – 3tan 3tan2Θ) =tan−1(3x3tanΘ– x3tan3Θx3 – 3x
= tan−1(tan3Θ)=3Θ=3tan−1ax
Q11. Solve tan−1[2cos(2sin −112)] Soln:
Given tan−1[2cos(2sin−112)]
tan−1[2cos(2sin −112)]=tan−1[2cos(2sin −1(sinπ6))] = tan−1[2cos(2×π6)]=tan −1[2cos(π3)]=tan −1[2×12] = tan−1[1]=π4
Q12. Solve cot(tan−1x+cot−1x) Soln:
Given cot(tan−1x+cot−1x)
cot(tan−1x+cot−1x)=cot(π2) =0
