Prove that sin^2(π/6) + cos^2(π/3) - tan^2(π/4) = -1/2.
Hint. Substitute the standard values for each of the three standard angles and simplify.
sin(pi/6)=1/2 so sin^2(pi/6)=1/4. cos(pi/3)=1/2 so cos^2(pi/3)=1/4. tan(pi/4)=1 so tan^2(pi/4)=1. Sum: 1/4+1/4-1 = 1/2-1 = -1/2.
✦ Working through each part gives: lHS = -1/2, matching the RHS.
Prove that 2sin^2(π/6) + cosec^2(7π/6)cos^2(π/3) = 3/2.
Hint. 7pi/6 = pi + pi/6, in the third quadrant -- find sin(7pi/6) using the allied-angle rule before squaring its reciprocal.
2sin^2(pi/6) = 2(1/2)^2 = 1/2. For 7pi/6 = pi+pi/6 (QIII): sin(7pi/6) = -sin(pi/6) = -1/2, so cosec(7pi/6) = -2, and cosec^2(7pi/6) = 4. cos^2(pi/3) = (1/2)^2 = 1/4. So the second term is 4 x 1/4 = 1. Total: 1/2 + 1 = 3/2.
✦ Working through each part gives: lHS = 3/2, matching the RHS.
Prove that cot^2(π/6) + cosec(5π/6) + 3tan^2(π/6) = 6.
Hint. 5pi/6 = pi - pi/6, in the second quadrant -- use the allied-angle rule for cosec(5pi/6).
cot(pi/6)=root3, so cot^2(pi/6)=3. For 5pi/6=pi-pi/6 (QII): sin(5pi/6)=sin(pi/6)=1/2, so cosec(5pi/6)=2. tan(pi/6)=1/root3, so tan^2(pi/6)=1/3, and 3tan^2(pi/6)=1. Total: 3+2+1=6.
✦ Working through each part gives: lHS = 6, matching the RHS.
Prove that 2sin^2(3π/4) + 2cos^2(π/4) + 2sec^2(π/3) = 10.
Hint. 3pi/4 = pi - pi/4; use the allied-angle rule to relate sin(3pi/4) to sin(pi/4).
sin(3pi/4)=sin(pi-pi/4)=sin(pi/4)=1/root2, so 2sin^2(3pi/4)=2(1/2)=1. cos(pi/4)=1/root2, so 2cos^2(pi/4)=2(1/2)=1. sec(pi/3)=1/cos(pi/3)=2, so 2sec^2(pi/3)=2(4)=8. Total: 1+1+8=10.
✦ Working through each part gives: lHS = 10, matching the RHS.
Find the value of: (i) sin 75° (ii) tan 15°.
Hint. Write 75deg = 45deg+30deg and 15deg = 45deg-30deg, then apply the sum and difference formulas.
(i) sin75deg = sin(45deg+30deg) = sin45cos30+cos45sin30 = (1/root2)(root3/2)+(1/root2)(1/2) = (root3+1)/(2root2). (ii) tan15deg = tan(45deg-30deg) = (tan45-tan30)/(1+tan45tan30) = (1-1/root3)/(1+1/root3) = (root3-1)/(root3+1), which rationalises to 2-root3.
✦ Working through each part gives: (i) sin75deg = (root3+1)/(2root2) (ii) tan15deg = 2-root3.
Prove that cos(π/4-x)cos(π/4-y) - sin(π/4-x)sin(π/4-y) = sin(x+y).
Hint. The left side has the exact shape of the cos(A+B) expansion -- identify A and B and collapse it in one step.
The LHS matches cosA cosB - sinA sinB = cos(A+B), with A=pi/4-x and B=pi/4-y. So LHS = cos[(pi/4-x)+(pi/4-y)] = cos(pi/2-(x+y)) = sin(x+y), using cos(pi/2-theta)=sin(theta).
✦ Working through each part gives: lHS = sin(x+y), matching the RHS.
Prove that tan(π/4+x) / tan(π/4-x) = [(1+tan x)/(1-tan x)]^2.
Hint. Expand tan(pi/4+x) and tan(pi/4-x) separately using tan(pi/4)=1, then divide.
tan(pi/4+x) = (1+tanx)/(1-tanx) and tan(pi/4-x) = (1-tanx)/(1+tanx), both using tan(pi/4)=1 in the tangent sum/difference formula. Dividing the first by the second: [(1+tanx)/(1-tanx)] / [(1-tanx)/(1+tanx)] = [(1+tanx)/(1-tanx)] x [(1+tanx)/(1-tanx)] = [(1+tanx)/(1-tanx)]^2.
✦ Working through each part gives: lHS = [(1+tanx)/(1-tanx)]^2, matching the RHS.
Prove that [cos(π+x)cos(-x)] / [sin(π-x)cos(π/2+x)] = cot^2 x.
Hint. Reduce each of the four allied-angle factors one at a time before dividing.
cos(pi+x)=-cosx and cos(-x)=cosx, so the numerator is -cosx.cosx=-cos^2x. sin(pi-x)=sinx and cos(pi/2+x)=-sinx, so the denominator is sinx.(-sinx)=-sin^2x. Dividing: (-cos^2x)/(-sin^2x) = cos^2x/sin^2x = cot^2x.
✦ Working through each part gives: lHS = cot^2 x, matching the RHS.
Prove that cos(3π/2+x)cos(2π+x)[cot(3π/2-x)+cot(2π+x)] = 1.
Hint. Reduce each of the four pieces separately: two are direct allied-angle or periodicity results, and cot(3pi/2-x) needs the co-function flip that happens at odd multiples of pi/2, just like at pi/2.
cos(3pi/2+x)=sinx. cos(2pi+x)=cosx (periodicity). For cot(3pi/2-x): sin(3pi/2-x)=-cosx and cos(3pi/2-x)=-sinx, so cot(3pi/2-x)=(-sinx)/(-cosx)=tanx. cot(2pi+x)=cotx (periodicity). So the bracket is tanx+cotx = sinx/cosx+cosx/sinx = (sin^2x+cos^2x)/(sinxcosx) = 1/(sinxcosx). Multiplying: sinx.cosx.[1/(sinxcosx)] = 1.
✦ Working through each part gives: lHS = 1, matching the RHS.
Where students slip. Assuming cot(3pi/2-x) reduces the same way as cot(pi-x) or cot(2pi-x) (same function, sign only) -- 3pi/2 is an odd multiple of pi/2, so it flips cot to tan, just as pi/2 does.
Prove that sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x = cos x.
Hint. The left side has the exact shape of the cos(A-B) expansion, with A=(n+1)x and B=(n+2)x.
cosA cosB + sinA sinB = cos(A-B), with A=(n+1)x and B=(n+2)x. So LHS = cos[(n+1)x-(n+2)x] = cos(-x) = cosx.
✦ Working through each part gives: lHS = cos x, matching the RHS.
Prove that cos(3π/4+x) - cos(3π/4-x) = -√2 sin x.
Hint. Use the sum-to-product formula for a difference of two cosines, with the average and half-difference angles simplifying immediately.
cosA-cosB = -2sin((A+B)/2)sin((A-B)/2), with A=3pi/4+x, B=3pi/4-x. (A+B)/2=3pi/4 and (A-B)/2=x. So LHS = -2sin(3pi/4)sinx = -2(1/root2)sinx = -root2 sinx.
✦ Working through each part gives: lHS = -root2 sin x, matching the RHS.
Prove that sin^2(6x) - sin^2(4x) = sin(2x) sin(10x).
Hint. Write both squared terms using sin^2(theta)=(1-cos2theta)/2 and subtract, then convert the resulting difference of cosines into a product.
sin^2(6x)-sin^2(4x) = [(1-cos12x)/2] - [(1-cos8x)/2] = (cos8x-cos12x)/2. Using cosA-cosB=-2sin((A+B)/2)sin((A-B)/2) with A=8x,B=12x: cos8x-cos12x = -2sin(10x)sin(-2x) = 2sin(10x)sin(2x). Dividing by 2 gives sin(2x)sin(10x).
✦ Working through each part gives: lHS = sin(2x)sin(10x), matching the RHS.
Prove that cos^2(2x) - cos^2(6x) = sin(4x) sin(8x).
Hint. Write both squared terms using cos^2(theta)=(1+cos2theta)/2 and subtract, then convert the resulting difference of cosines into a product.
cos^2(2x)-cos^2(6x) = [(1+cos4x)/2]-[(1+cos12x)/2] = (cos4x-cos12x)/2. Using cosA-cosB=-2sin((A+B)/2)sin((A-B)/2) with A=4x,B=12x: cos4x-cos12x = -2sin(8x)sin(-4x) = 2sin(8x)sin(4x). Dividing by 2 gives sin(4x)sin(8x).
✦ Working through each part gives: lHS = sin(4x)sin(8x), matching the RHS.
Prove that sin^2 x + 2sin(4x) + sin(6x) = 4cos^2 x sin(4x).
Hint. Group sin2x and sin6x with the sum-to-product formula, then factor out the common sin4x term.
sin2x+sin6x = 2sin4x cos2x. Adding the remaining sin4x term: LHS = 2sin4x cos2x + 2sin4x = 2sin4x(cos2x+1). Using 1+cos2x=2cos^2x: 2sin4x x 2cos^2x = 4cos^2x sin4x.
✦ Working through each part gives: lHS = 4cos^2 x sin4x, matching the RHS.
Where students slip. Misreading the question's first term as sin(2x) instead of sin^2 x -- the identity only balances when the first term is sin2x (i.e. this is really sin2x + 2sin4x + sin6x, matching the NCERT original).
Prove that cot(4x)(sin5x + sin3x) = cot(x)(sin5x - sin3x).
Hint. Convert both bracketed sums to products first, then simplify each side of the equation independently and compare.
sin5x+sin3x = 2sin4x cosx, and sin5x-sin3x = 2cos4x sinx. LHS = cot4x x 2sin4x cosx = (cos4x/sin4x) x 2sin4x cosx = 2cos4x cosx. RHS = cotx x 2cos4x sinx = (cosx/sinx) x 2cos4x sinx = 2cos4x cosx. Both sides equal 2cos4x cosx.
✦ Working through each part gives: both sides equal 2cos(4x)cos(x), so the identity holds.
Prove that (cos9x - cos5x) / (sin17x - sin3x) = -sin2x / cos10x.
Hint. Convert the numerator and denominator to products separately using the sum-to-product formulas, then cancel the shared factor.
cos9x-cos5x = -2sin7x sin2x. sin17x-sin3x = 2cos10x sin7x. Dividing: (-2sin7x sin2x)/(2cos10x sin7x) = -sin2x/cos10x, cancelling the shared sin7x factor.
✦ Working through each part gives: lHS = -sin2x/cos10x, matching the RHS.
Prove that (sin5x + sin3x) / (cos5x + cos3x) = tan4x.
Hint. Convert both the numerator and denominator to products; a common factor cancels, leaving a single ratio of sine to cosine.
sin5x+sin3x = 2sin4x cosx. cos5x+cos3x = 2cos4x cosx. Dividing cancels the shared 2cosx factor, leaving sin4x/cos4x = tan4x.
✦ Working through each part gives: lHS = tan4x, matching the RHS.
Prove that (sin x - sin y) / (cos x + cos y) = tan((x-y)/2).
Hint. Convert numerator and denominator to products; the (x+y)/2 factors are common to both and cancel.
sinx-siny = 2cos((x+y)/2)sin((x-y)/2). cosx+cosy = 2cos((x+y)/2)cos((x-y)/2). Dividing cancels the shared 2cos((x+y)/2) factor, leaving sin((x-y)/2)/cos((x-y)/2) = tan((x-y)/2).
✦ Working through each part gives: lHS = tan((x-y)/2), matching the RHS.
Prove that (sin x + sin 3x) / (cos x + cos 3x) = tan 2x.
Hint. Convert both numerator and denominator to products with x and 3x playing the roles of the two angles.
sinx+sin3x = 2sin2x cosx. cosx+cos3x = 2cos2x cosx. Dividing cancels the shared 2cosx factor, leaving sin2x/cos2x = tan2x.
✦ Working through each part gives: lHS = tan2x, matching the RHS.
Prove that (sin x - sin y) / (cos x - cos y) = -cot((x+y)/2).
Hint. Convert numerator and denominator to products; notice the denominator picks up a negative sign that the numerator does not.
sinx-siny = 2cos((x+y)/2)sin((x-y)/2). cosx-cosy = -2sin((x+y)/2)sin((x-y)/2). Dividing cancels the shared sin((x-y)/2) factor: [2cos((x+y)/2)]/[-2sin((x+y)/2)] = -cos((x+y)/2)/sin((x+y)/2) = -cot((x+y)/2).
✦ Working through each part gives: lHS = -cot((x+y)/2), matching the RHS.
Where students slip. Dropping the negative sign that comes from cosx-cosy = -2sin(..)sin(..) -- unlike the other three sum-to-product formulas, this is the one with a leading minus sign.
Prove that (cos4x + cos3x + cos2x) / (sin4x + sin3x + sin2x) = cot 3x.
Hint. Pair the two outer terms (cos4x with cos2x, sin4x with sin2x) around the middle term first, then factor.
cos4x+cos2x = 2cos3x cosx; adding cos3x gives cos3x(2cosx+1). Similarly sin4x+sin2x = 2sin3x cosx; adding sin3x gives sin3x(2cosx+1). Dividing cancels the shared (2cosx+1) factor, leaving cos3x/sin3x = cot3x.
✦ Working through each part gives: lHS = cot3x, matching the RHS.
Prove that cot x cot 2x - cot 2x cot 3x - cot 3x cot x = 1.
Hint. Write 3x = x+2x and expand cot(3x) using the cotangent sum formula, then rearrange the resulting equation.
Since 3x=x+2x: cot3x = (cotx.cot2x - 1)/(cotx+cot2x), so cot3x(cotx+cot2x) = cotx.cot2x - 1. Expanding: cot3x.cotx + cot3x.cot2x = cotx.cot2x - 1. Rearranging: cotx.cot2x - cot2x.cot3x - cot3x.cotx = 1.
✦ Working through each part gives: lHS = 1, matching the RHS after rearranging the cot(x+y) expansion.
Prove that tan 4x = [4tan x(1 - tan^2 x)] / [1 - 6tan^2 x + tan^4 x].
Hint. Apply the double-angle formula for tangent twice: first get tan2x in terms of tanx, then tan4x in terms of tan2x, and simplify the resulting compound fraction.
Let t=tanx. tan2x = 2t/(1-t^2). Then tan4x = 2tan2x/(1-tan^2(2x)) = [4t/(1-t^2)] / [1 - 4t^2/(1-t^2)^2] = [4t/(1-t^2)] x [(1-t^2)^2/((1-t^2)^2-4t^2)] = 4t(1-t^2)/((1-t^2)^2-4t^2). Expanding the denominator: (1-t^2)^2-4t^2 = 1-2t^2+t^4-4t^2 = 1-6t^2+t^4.
✦ Working through each part gives: tan4x = 4tanx(1-tan^2x) / (1-6tan^2x+tan^4x), matching the RHS.
Prove that cos 4x = 1 - 8sin^2 x cos^2 x.
Hint. Write cos4x using the double-angle formula with argument 2x, then substitute sin2x=2sinxcosx.
cos4x = cos(2x2x) = 1-2sin^2(2x), using cos2theta=1-2sin^2theta with theta=2x. Since sin2x=2sinxcosx, sin^2(2x)=4sin^2xcos^2x. So cos4x = 1-2(4sin^2xcos^2x) = 1-8sin^2xcos^2x.
✦ Working through each part gives: lHS = 1-8sin^2xcos^2x, matching the RHS.
Prove that cos 6x = 32cos^6 x - 48cos^4 x + 18cos^2 x - 1.
Hint. Write cos6x as the triple-angle formula applied to 2x, then substitute cos2x=2cos^2x-1 and expand carefully.
cos6x = cos(3.2x) = 4cos^3(2x) - 3cos(2x), the triple-angle formula with argument 2x. Let c=cosx, so cos2x=2c^2-1. Expanding (2c^2-1)^3 = 8c^6-12c^4+6c^2-1, so 4cos^3(2x) = 32c^6-48c^4+24c^2-4. And 3cos2x = 6c^2-3. Subtracting: cos6x = (32c^6-48c^4+24c^2-4) - (6c^2-3) = 32c^6-48c^4+18c^2-1.
✦ Working through each part gives: cos6x = 32cos^6x - 48cos^4x + 18cos^2x - 1, matching the RHS.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh103.pdf) — three numbered exercises (3.1-3.3, 42 questions) plus the chapter's Miscellaneous Exercise (10 questions); the chapter has no Exercise 3.4 and no general-solution-of-trigonometric-equations content, contrary to an earlier stub. Miscellaneous Exercise Q1's stacked-fraction identity was re-rendered directly from the PDF at 300dpi after raw text extraction garbled the numerators, confirming it as 2cos(pi/13)cos(9pi/13)+cos(3pi/13)+cos(5pi/13)=0. Questions are referenced from the NCERT textbook for identification.
Pair self-study with a tutor, a live course or a coaching centre.
