NCERT Exemplar Class 10 Maths Chapter 8 Exercise 8.1 Solutions

Turn one ratio into a triangle, then read the rest.

• Build sides: cos A compares adjacent to hypotenuse, so 4 and 5 are two sides of a right triangle.
• Third side: the missing leg has to be 3, the familiar Pythagorean partner of 4 and 5.
• Read tangent: with all three sides known, tangent pairs the opposite 3 with the adjacent 4, giving 34.
• Common slip: students who answer 45 forget that tangent never uses the hypotenuse.

Option (B), 34.

Name the angle first.

• Spot the angle: a sine of 12 points straight to 30^∘, and the whole standard-angle row then unlocks.
• Read cotangent: cot 30^∘=3, the largest of the three table cotangents for the common angles.
• Check with sides: opposite 1, adjacent 3, so cot A=3 again.
• Trap option: 13 is tan 30^∘, the reciprocal that students grab when they swap opposite and adjacent.

Option (A), 3.

Two cancelling couples, nothing left.

• First couple: csc(75^∘+θ) and sec(15^∘-θ) are equal because their angles are complements, so their difference is zero whatever θ is.
• Second couple: tan(55^∘+θ) and cot(35^∘-θ) are equal for the same reason, and the signs make them cancel too.
• Why it is flat: because θ disappears from both couples, the answer is a flat 0 for every allowed value of θ.

Option (B), 0.

Picture the triangle behind the fraction.

• Label sides: if sinθ=ab, take opposite =a and hypotenuse =b.
• Find adjacent: by Pythagoras the adjacent side is √b²-a², and cosine compares it to the hypotenuse b.
• Read the answer: so cosθ=√b²-a²b falls out without algebra.
• Cross-check: the identity route gives the same thing, and both keep the hypotenuse b in the denominator.

Option (C), √b²-a²b.

One equation removes one unknown.

• Pin the sum: the clue cos(α+β)=0 forces the sum to 90^∘, so α and β are complements.
• Rewrite the gap: that lets you write α-β purely in β, where it becomes 90^∘-2β.
• Apply the rule: the sine of that angle, by the complementary rule, is cos 2β.
• Sanity filter: the answer keeps only β, which rules out the α-only options at a glance.

Option (B), cos 2β.

Symmetry around 45^∘ does all the work.

• Count the terms: the factors run from 1^∘ to 89^∘, an odd count of 89 terms symmetric about 45^∘.
• Pair the outsides: each outside pair multiplies a tangent by the cotangent of the same small angle, which is 1.
• Lone middle term: after all 44 pairs vanish to 1, only tan 45^∘=1 is left in the centre.
• No growth: so the giant product is simply 1, not something that grows with the number of terms.

Option (B), 1.

Convert, equate, solve.

• The hurdle: one side is a cosine and the other a sine, so they cannot be compared as they stand.
• Convert: the complementary rule rewrites sinα as cos(90^∘-α), so the equation reads cosine equals cosine.
• Equate and solve: with both angles acute they must be identical, giving 9α=90^∘-α and α=9^∘.
• Finish: then 5α=45^∘, which lands on the friendliest standard value, 1.

Option (C), 1.

The right angle fixes the other two.

• Use the angle sum: a right angle at C takes up 90^∘ of the triangle's 180^∘, leaving exactly 90^∘ for A and B.
• Locked sum: so A+B stays at 90^∘ no matter how the triangle is shaped, and the cosine of 90^∘ is 0.
• Shortcut: there is no need to know A or B separately, since only their sum matters here.

Option (A), 0.

One substitution links everything.

• The key move: read sin A+sin² A=1 as sin A=cos² A, because 1-sin² A is exactly cos² A.
• Swap the target: with that bridge the target's cos² A becomes sin A and its cos⁴ A becomes sin² A.
• Circle back: so cos² A+cos⁴ A rebuilds sin A+sin² A, which the problem already says is 1.
• By design: the expression was engineered to circle straight back to the given condition.

Option (A), 1.

Two table look-ups, then add.

• Read sine: a sine of 12 means α=30^∘ from the standard table.
• Read cosine: a cosine of 12 means β=60^∘.
• Add them: these are complementary angles, so their sum returns 90^∘.
• Shortcut: since cosβ=12=sinα forces β=90^∘-α, the sum is 90^∘ at once.

Option (D), 90^∘.

Three hidden ones, then 1+1.

• Numerator: the complement pair rebuilds sin² 22^∘+cos² 22^∘=1.
• Denominator: the same trick rebuilds another 1, so the whole fraction is 1.
• The tail: sin 27^∘ is just cos 63^∘, turning the product into cos² 63^∘, which joins sin² 63^∘ to make a third 1.
• Add up: adding the ones gives 2, and nothing depends on the decimal value of any angle.

Option (B), 2.

One division removes the angle.

• Same degree: both top and bottom are first-degree in sinθ and cosθ, so dividing by cosθ converts everything to tanθ alone.
• Slot the given: the given 4tanθ=3 then makes the numerator 3-1=2 and the denominator 3+1=4.
• Reduce: the ratio 24 reduces to 12, with no need to find θ or the individual ratios.

Option (C), 12.

Pin the angle, then plug in.

• Read the condition: it is really tanθ=1, which singles out 45^∘ in the acute range.
• Plug in: there both sine and cosine equal 12, so each fourth power is 14 and the sum is 12.
• Algebraic check: using sin⁴θ+cos⁴θ=1-2sin²²θ with sin²θ=cos²θ=12 gives the same 12.

Option (C), 12.

The two angles are complements, so the terms match.

• Add the angles: (45^∘+θ) and (45^∘-θ) sum to 90^∘ as the θ terms cancel.
• Match the terms: that makes cos(45^∘-θ) the same as sin(45^∘+θ).
• Subtract: the expression then subtracts a quantity from itself, so the result is 0 for every θ.
• Spot the decoys: the tempting answers involving θ are traps, since the symmetry wipes θ out entirely.

Option (B), 0.