Find the principal value of sin^{-1}(-1/2).
Hint. Find the angle in [-pi/2,pi/2] whose sine is -1/2.
Since sin(pi/6)=1/2, sin(-pi/6)=-1/2, and -pi/6 lies in the principal branch [-pi/2,pi/2], because sine is an odd function so its inverse's principal value inherits the sign of the input.
✦ -pi/6
Find the principal value of cos^{-1}(sqrt3/2).
Hint. Find the angle in [0,pi] whose cosine is sqrt3/2.
Since cos(pi/6)=sqrt3/2, and pi/6 lies in the principal branch [0,pi], because this is a direct standard-angle match with no sign adjustment needed.
✦ pi/6
Find the principal value of cosec^{-1}(2).
Hint. Find the angle in [-pi/2,pi/2]-{0} whose cosecant is 2.
Since sin(pi/6)=1/2, cosec(pi/6)=2, and pi/6 lies in the principal branch [-pi/2,pi/2]-{0}, because this is a direct standard-angle match.
✦ pi/6
Find the principal value of tan^{-1}(-sqrt3).
Hint. Find the angle in (-pi/2,pi/2) whose tangent is -sqrt3.
Since tan(pi/3)=sqrt3, tan(-pi/3)=-sqrt3, and -pi/3 lies in the principal branch (-pi/2,pi/2), because tangent is odd so the sign carries directly to the principal value.
✦ -pi/3
Find the principal value of cos^{-1}(-1/2).
Hint. Find the angle in [0,pi] whose cosine is -1/2, using the supplementary-angle relation for negative cosine values.
Since cos(pi/3)=1/2, and cosine of a negative value corresponds to the supplement, cos(pi-pi/3)=cos(2pi/3)=-1/2, and 2pi/3 lies in the principal branch [0,pi], because the negative sign shifts the angle to the supplement rather than a negative angle (unlike sin^{-1} and tan^{-1}).
✦ 2pi/3
Find the principal value of tan^{-1}(-1).
Hint. Find the angle in (-pi/2,pi/2) whose tangent is -1.
Since tan(pi/4)=1, tan(-pi/4)=-1, and -pi/4 lies in the principal branch (-pi/2,pi/2), because tangent is odd.
✦ -pi/4
Find the principal value of sec^{-1}(2/sqrt3).
Hint. Find the angle in [0,pi]-{pi/2} whose secant is 2/sqrt3.
Since cos(pi/6)=sqrt3/2, sec(pi/6)=2/sqrt3, and pi/6 lies in the principal branch [0,pi]-{pi/2}, because this is a direct standard-angle match.
✦ pi/6
Find the principal value of cot^{-1}(sqrt3).
Hint. Find the angle in (0,pi) whose cotangent is sqrt3.
Since cot(pi/6)=sqrt3, and pi/6 lies in the principal branch (0,pi), because this is a direct standard-angle match.
✦ pi/6
Find the principal value of cos^{-1}(-1/sqrt2).
Hint. Use the supplementary-angle relation for a negative cosine value.
Since cos(pi/4)=1/sqrt2, cos(pi-pi/4)=cos(3pi/4)=-1/sqrt2, and 3pi/4 lies in the principal branch [0,pi], because negative cosine values map to the supplement of the reference angle.
✦ 3pi/4
Find the principal value of cosec^{-1}(-sqrt2).
Hint. Find the angle in [-pi/2,pi/2]-{0} whose cosecant is -sqrt2.
Since sin(pi/4)=1/sqrt2, cosec(pi/4)=sqrt2, so cosec(-pi/4)=-sqrt2, and -pi/4 lies in the principal branch [-pi/2,pi/2]-{0}, because cosecant is odd so the sign carries directly.
✦ -pi/4
Find the value of tan^{-1}(1) + cos^{-1}(-1/2) + sin^{-1}(-1/2).
Hint. Evaluate each of the three principal values separately, then add using a common denominator.
tan^{-1}(1)=pi/4. cos^{-1}(-1/2)=2pi/3 (supplement of pi/3, since cosine of a negative value maps to the supplement). sin^{-1}(-1/2)=-pi/6 (odd function). Using twelfths: pi/4=3pi/12, 2pi/3=8pi/12, -pi/6=-2pi/12, so the sum is (3+8-2)pi/12=9pi/12=3pi/4.
✦ 3pi/4
Find the value of cos^{-1}(1/2) + 2sin^{-1}(1/2).
Hint. Evaluate each principal value first, then combine.
cos^{-1}(1/2)=pi/3. sin^{-1}(1/2)=pi/6, so 2sin^{-1}(1/2)=pi/3. Adding: pi/3+pi/3=2pi/3.
✦ 2pi/3
If sin^{-1}x=y, then (A) 0<=y<=pi (B) -pi/2<=y<=pi/2 (C) 0<y<pi (D) -pi/2<y<pi/2.
Hint. Recall the principal value branch (a closed interval) of sin^{-1}.
By definition, the principal value branch of sin^{-1} is the closed interval [-pi/2,pi/2], since the endpoints pi/2 and -pi/2 are themselves valid principal values (e.g. sin^{-1}(1)=pi/2).
✦ (B) -pi/2<=y<=pi/2
tan^{-1}(sqrt3) - sec^{-1}(-2) is equal to (A) pi (B) -pi/3 (C) pi/3 (D) 2pi/3.
Hint. Evaluate each principal value separately, being careful with the sign convention for sec^{-1} of a negative number.
tan^{-1}(sqrt3)=pi/3. For sec^{-1}(-2): need cos(theta)=-1/2 with theta in [0,pi]-{pi/2}, which gives theta=2pi/3 (the supplement of pi/3, following the same negative-value convention as cos^{-1}). So the expression is pi/3-2pi/3=-pi/3.
✦ (B) -pi/3
Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh102.pdf) — Exercise 2.1 (14 questions), Exercise 2.2 (15 questions), plus the chapter's Miscellaneous Exercise (14 questions), 43 questions total. Exercise pages were rendered as 300dpi images to read the stacked-fraction inverse-trig notation accurately, since raw text extraction badly garbled it. The old stub had no solutions file at all and collapsed the three real exercises into one invented 21-question group; it also taught a numbered list of 'properties' (like sin^{-1}x+cos^{-1}x=pi/2) that the current edition does not box anywhere — confirmed by reading the full text of Section 2.3, which goes directly from a one-line recap into three worked examples that all use the same sinθ/cosθ/tanθ/secθ substitution technique rather than stating identities to memorise. Every proof-based answer in this file was independently re-derived from the substitution technique (not copied from a key), and Miscellaneous Q14 was specifically checked for the extraneous root x=1/2 that the squaring-equivalent step introduces but that fails the original equation.. Questions are referenced from the NCERT textbook for identification.
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