Find the value of cos^{-1}(cos(13pi/6)).
Hint. 13pi/6 is more than a full rotation past a standard angle — reduce it first using the period of cosine.
13pi/6=2pi+pi/6, and cosine has period 2pi, so cos(13pi/6)=cos(pi/6)=sqrt3/2. Since pi/6 is inside [0,pi], cos^{-1}(cos(13pi/6))=pi/6.
✦ pi/6
Find the value of tan^{-1}(tan(7pi/6)).
Hint. Use the period-pi property of tangent to reduce 7pi/6 to an angle inside the principal branch.
7pi/6=pi+pi/6, and tangent has period pi, so tan(7pi/6)=tan(pi/6)=1/sqrt3. Since pi/6 is inside (-pi/2,pi/2), tan^{-1}(tan(7pi/6))=pi/6.
✦ pi/6
Prove that 2sin^{-1}(3/5) = tan^{-1}(24/7).
Hint. Let alpha=sin^{-1}(3/5), extract its tangent ratio, then apply the double-angle tangent formula.
Let alpha=sin^{-1}(3/5): sin(alpha)=3/5, cos(alpha)=4/5, tan(alpha)=3/4, with alpha in (0,pi/2). tan(2alpha)=2tan(alpha)/(1-tan^2(alpha))=2(3/4)/(1-9/16)=(3/2)/(7/16)=24/7. Since alpha is about 37 degrees, 2alpha is about 74 degrees, which is inside (-pi/2,pi/2), so 2alpha=tan^{-1}(24/7). Therefore 2sin^{-1}(3/5)=tan^{-1}(24/7).
✦ Identity proved: 2sin^{-1}(3/5) = tan^{-1}(24/7).
Prove that sin^{-1}(8/17) + sin^{-1}(3/5) = tan^{-1}(77/36).
Hint. Extract the tangent ratio of each angle from a right triangle, then apply the tangent addition formula.
Let alpha=sin^{-1}(8/17): tan(alpha)=8/15. Let beta=sin^{-1}(3/5): tan(beta)=3/4, both alpha and beta in (0,pi/2). tan(alpha+beta) = (8/15+3/4)/(1-(8/15)(3/4)) = (77/60)/(3/5) = 77/36. Since alpha is about 28 degrees and beta about 37 degrees, their sum is under 90 degrees, inside the principal branch of tan^{-1}. Therefore sin^{-1}(8/17)+sin^{-1}(3/5)=tan^{-1}(77/36).
✦ Identity proved: sin^{-1}(8/17) + sin^{-1}(3/5) = tan^{-1}(77/36).
Prove that cos^{-1}(4/5) + cos^{-1}(12/13) = cos^{-1}(33/65).
Hint. Extract sine and cosine of each angle, then use the cosine addition formula.
Let alpha=cos^{-1}(4/5): sin(alpha)=3/5. Let beta=cos^{-1}(12/13): sin(beta)=5/13, both alpha and beta in (0,pi/2). cos(alpha+beta) = cos(alpha)cos(beta)-sin(alpha)sin(beta) = (4/5)(12/13)-(3/5)(5/13) = 48/65-15/65 = 33/65. Since alpha and beta are both in (0,pi/2), their sum is in (0,pi), matching the principal branch of cos^{-1}. Therefore cos^{-1}(4/5)+cos^{-1}(12/13)=cos^{-1}(33/65).
✦ Identity proved: cos^{-1}(4/5) + cos^{-1}(12/13) = cos^{-1}(33/65).
Prove that cos^{-1}(12/13) + sin^{-1}(3/5) = sin^{-1}(56/65).
Hint. Extract sine and cosine of each angle, then use the sine addition formula.
Let alpha=cos^{-1}(12/13): sin(alpha)=5/13, cos(alpha)=12/13. Let beta=sin^{-1}(3/5): sin(beta)=3/5, cos(beta)=4/5, both alpha and beta in (0,pi/2). sin(alpha+beta) = sin(alpha)cos(beta)+cos(alpha)sin(beta) = (5/13)(4/5)+(12/13)(3/5) = 20/65+36/65 = 56/65. Since alpha is about 23 degrees and beta about 37 degrees, their sum is under 90 degrees, inside the principal branch of sin^{-1}. Therefore cos^{-1}(12/13)+sin^{-1}(3/5)=sin^{-1}(56/65).
✦ Identity proved: cos^{-1}(12/13) + sin^{-1}(3/5) = sin^{-1}(56/65).
Prove that tan^{-1}(63/16) = sin^{-1}(5/13) + cos^{-1}(3/5).
Hint. Extract the tangent ratio of each angle on the right side, then apply the tangent addition formula.
Let alpha=sin^{-1}(5/13): tan(alpha)=5/12. Let beta=cos^{-1}(3/5): tan(beta)=4/3, both alpha and beta in (0,pi/2). tan(alpha+beta) = (5/12+4/3)/(1-(5/12)(4/3)) = (7/4)/(4/9) = 63/16. Since alpha is about 23 degrees and beta about 53 degrees, their sum is under 90 degrees, inside the principal branch of tan^{-1}. Therefore sin^{-1}(5/13)+cos^{-1}(3/5)=tan^{-1}(63/16), which is the required identity written the other way round.
✦ Identity proved: tan^{-1}(63/16) = sin^{-1}(5/13) + cos^{-1}(3/5).
Prove that tan^{-1}(sqrt x) = (1/2)cos^{-1}((1-x)/(1+x)), x in [0,1].
Hint. Substitute x=tan^2θ, then recognise the right side as a cosine double-angle formula.
Let x=tan^2θ with θ in [0,pi/4] (since x is in [0,1]), so sqrt x=tanθ. The left side becomes tan^{-1}(tanθ)=θ. For the right side: (1-tan^2θ)/(1+tan^2θ)=cos2θ, a standard identity, so (1/2)cos^{-1}(cos2θ)=(1/2)(2θ)=θ, valid since 2θ is in [0,pi/2], inside the principal branch [0,pi] of cos^{-1}. Both sides equal θ.
✦ Identity proved: tan^{-1}(sqrt x) = (1/2)cos^{-1}((1-x)/(1+x)) for x in [0,1].
Prove that cot^{-1}[(sqrt(1+sin x)+sqrt(1-sin x))/(sqrt(1+sin x)-sqrt(1-sin x))] = x/2, x in (0,pi/4).
Hint. Write 1+sin x and 1-sin x each as a perfect square in terms of sin(x/2) and cos(x/2).
Since 1+sin x = (cos(x/2)+sin(x/2))^2 and 1-sin x = (cos(x/2)-sin(x/2))^2, and for x in (0,pi/4), x/2 is in (0,pi/8), where cos(x/2)>sin(x/2)>0, so sqrt(1+sin x)=cos(x/2)+sin(x/2) and sqrt(1-sin x)=cos(x/2)-sin(x/2). The numerator is then 2cos(x/2) and the denominator is 2sin(x/2), giving the ratio cot(x/2). So the expression is cot^{-1}(cot(x/2))=x/2, valid since x/2 is inside the principal branch (0,pi) of cot^{-1}.
✦ Identity proved: the expression equals x/2 for x in (0,pi/4).
Prove that tan^{-1}[(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))] = pi/4 - (1/2)cos^{-1}x, -1/sqrt2<=x<=1. [Hint: put x=cos2θ]
Hint. With x=cos2θ, rewrite 1+x and 1-x using the cosine double-angle formula, then simplify the ratio to a tangent-subtraction form.
Let x=cos2θ with θ in [0,3pi/8] (matching the given range of x). Then 1+x=2cos^2θ and 1-x=2sin^2θ, so sqrt(1+x)=sqrt2 cosθ and sqrt(1-x)=sqrt2 sinθ (both cosθ,sinθ non-negative on this interval). The ratio becomes (cosθ-sinθ)/(cosθ+sinθ) = (1-tanθ)/(1+tanθ) = tan(pi/4-θ), using the tangent subtraction formula. So the left side is tan^{-1}(tan(pi/4-θ))=pi/4-θ, valid since pi/4-θ stays inside (-pi/2,pi/2) for this range of θ. Since x=cos2θ with 2θ in [0,pi], θ=(1/2)cos^{-1}x. Substituting gives pi/4-(1/2)cos^{-1}x, matching the right side.
✦ Identity proved: the expression equals pi/4 - (1/2)cos^{-1}x for the given range of x.
Solve 2tan^{-1}(cos x) = tan^{-1}(2 cosec x).
Hint. Apply the double-angle formula for tan^{-1} to the left side, then equate the arguments (since tan^{-1} is one-one).
2tan^{-1}(cos x) = tan^{-1}(2cos x/(1-cos^2 x)) = tan^{-1}(2cos x/sin^2 x), using the double-angle tan^{-1} formula. Equating arguments with the right side: 2cos x/sin^2 x = 2/sin x. Multiplying both sides by sin^2 x (nonzero, since cosec x must be defined): 2cos x = 2sin x, so cos x = sin x, giving tan x=1, so x=pi/4 (the principal solution). Verification: at x=pi/4, both sides evaluate to approximately 70.53 degrees, confirming the solution.
✦ x = pi/4
Solve tan^{-1}((1-x)/(1+x)) = (1/2)tan^{-1}x, (x>0).
Hint. Recognise the left side as the standard tan^{-1}(1) - tan^{-1}x pattern.
With x=tan(alpha), (1-x)/(1+x) = (tan(pi/4)-tan(alpha))/(1+tan(pi/4)tan(alpha)) = tan(pi/4-alpha), so the left side is pi/4-tan^{-1}x. The equation becomes pi/4-tan^{-1}x = (1/2)tan^{-1}x, so pi/4 = (3/2)tan^{-1}x, giving tan^{-1}x=pi/6, so x=tan(pi/6)=1/sqrt3. Since x>0 is required and 1/sqrt3>0, this solution is valid.
✦ x = 1/sqrt3
sin(tan^{-1}x), |x|<1 is equal to (A) x/sqrt(1-x^2) (B) 1/sqrt(1-x^2) (C) 1/sqrt(1+x^2) (D) x/sqrt(1+x^2).
Hint. Let tan^{-1}x=theta, build a right triangle with opposite x and adjacent 1, then read off sin(theta).
Let theta=tan^{-1}x, so tan(theta)=x, with theta in (-pi/4,pi/4) roughly since |x|<1. In the corresponding right triangle, opposite=x, adjacent=1, hypotenuse=sqrt(1+x^2). So sin(theta)=x/sqrt(1+x^2), with the sign automatically matching the sign of x since theta is in (-pi/2,pi/2).
✦ (D) x/sqrt(1+x^2)
If sin^{-1}(1-x) - 2sin^{-1}x = pi/2, then x is equal to (A) 0, 1/2 (B) 1, 1/2 (C) 0 (D) 1/2.
Hint. Let sin^{-1}x=alpha, rewrite the equation using cos(2alpha)=1-2sin^2(alpha), solve the resulting quadratic, then check both roots against the original equation.
Let sin^{-1}x=alpha, so x=sin(alpha). The equation becomes sin^{-1}(1-x)=pi/2+2alpha, so 1-x=sin(pi/2+2alpha)=cos(2alpha)=1-2x^2. This gives -x=-2x^2, so 2x^2-x=0, so x(2x-1)=0, giving x=0 or x=1/2. Checking x=1/2 in the original equation: sin^{-1}(1/2)-2sin^{-1}(1/2)=pi/6-pi/3=-pi/6, which does not equal pi/2 — extraneous. Checking x=0: sin^{-1}(1)-2sin^{-1}(0)=pi/2-0=pi/2 — valid. So only x=0 actually satisfies the original equation.
✦ (C) 0
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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