NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry

How to present this for full marks.

• Hypotenuse first: a board examiner wants the longest side worked out before any ratio, so state Pythagoras once to get AC=√24²+7²=√625=25 cm, and spotting the standard (7,24,25) triple confirms that value at a single glance.
• Name the sides: before you substitute, write down which side is opposite the angle and which is beside it, since for angle A the opposite side is BC=7 and the adjacent side is AB=24, so the two ratios follow at once.
• Swap for C: the same two sides switch their roles at the other corner of the triangle, and naming them again keeps the working clear, giving sin C=2425 and cos C=725.
• Why it scores: writing the named side beside each fraction earns the method marks even if a final arithmetic slip creeps in, and it makes the pattern that links the two angles an easy self-check at the end.

sin A=cos C=725 and cos A=sin C=2425, with AC=25 cm.

Spot the Pythagorean triple.

• Read the triple: with PQ=12 and PR=13 the missing leg is QR=5, so the longer square-root step is skipped and a sign slip avoided.
• Tangent at P: this is the side facing P over the side beside it, QRPQ=512.
• Cotangent at R: the side beside R over the side facing it gives the same 512, because P and R are complementary, so their difference vanishes.
• Worth memorising: keeping the common triples (3,4,5), (5,12,13) and (8,15,17) in mind lets you head straight to the ratios, a real time saver in the board exam.

tan P=cot R=512, so their difference is 0.

Leave the surd as it is.

• Build the triangle: from sin A=34 set opposite =3 and hypotenuse =4, so the adjacent side is √4²-3²=7, an irrational number.
• Read cosine: the adjacent over the hypotenuse gives cos A=74 directly, with no rounding needed.
• Two equal forms: for the tangent both 37 and the rationalised 377 are correct and equal, so either one earns full marks.
• Stay exact: keeping the root is the safest choice in a non-calculator paper, since a decimal such as 1.134 would only be an approximation and could be marked down.

cos A=74, tan A=37=377.

Use the (8,15,17) triple.

• Fix the legs: the given relation gives cot A=815, so the adjacent side is 8 and the opposite side is 15.
• Hypotenuse free: the legs 8 and 15 belong to the standard triple, so the hypotenuse is 17 at sight, with no square-root step and no chance of error.
• Read the answers: straight off the triple, sin A=1517 and sec A=178.
• Speed habit: memorising the four common triples turns these ``one ratio given, find the rest'' questions into two-line answers, exactly the speed the board exam rewards.

sin A=1517, sec A=178, using the (8,15,17) triple.

Build a reference triangle once.

• Sketch the triple: draw a right triangle with adjacent 12, opposite 5 and hypotenuse 13, the familiar (5,12,13) shape.
• Sine and cosine: both come from the hypotenuse, giving sinθ=513 and cosθ=1213.
• Tangent and reciprocals: opposite over adjacent is 512, and flipping the ratios gives cotθ=125 and cscθ=135.
• Why draw it: one labelled triangle replaces six separate definitions, so the ratios fall out together in a single pass with no repeated arithmetic.

sinθ=513, cosθ=1213, tanθ=512, cotθ=125, cscθ=135.

A clean proof states the converse plainly.

• Cancel the hypotenuse: both A and B share the hypotenuse AB, so the given cos A=cos B reduces directly to the equal sides AC=BC.
• Quote the theorem: equal sides of a triangle lie opposite equal angles; this is the standard converse of the isosceles-triangle property, and naming it is what completes the argument.
• Read the angles: the side opposite A is BC and the side opposite B is AC, so the equal sides force ∠ A=∠ B.
• Where the marks sit: the credit here is for the reasoning, not the figure, so always state the property you rely on rather than letting the equal sides speak for themselves.

cos A=cos B⇒ AC=BC⇒∠ A=∠ B.

Recognise the structure before substituting.

• Numerator: the product (1+sinθ)(1-sinθ) is a difference of squares equal to 1-sin²θ=cos²θ.
• Denominator: likewise (1+cosθ)(1-cosθ) collapses to 1-cos²θ=sin²θ.
• One quantity: their ratio is exactly cot²θ, so both parts of the question are the same value (78)²=4964 at once.
• General habit: when a question gives one ratio and asks for an expression, simplify it symbolically first; very often it collapses to a power of the ratio you were handed, as here, which saves all the side-by-side arithmetic of a triangle approach.

Both parts equal cot²θ=4964.

This is a genuine identity, not a coincidence.

• Clear the tangents: multiply the top and bottom of the left side by cos² A to reach cos² A-sin² Acos² A+sin² A.
• Use Pythagoras: the new denominator is cos² A+sin² A=1, so the left side becomes cos² A-sin² A, which is exactly the right side.
• True for all A: the relation therefore holds for every acute angle, and the matching value 725 here is just one instance of it.
• Safe in the exam: recognising it as an identity lets you answer ``yes'' even without the angle, but showing the equal value on both sides is the surest way to earn full marks in a ``check whether'' question.

The equation is a true identity; here both sides are 725.

See the angle-sum pattern.

• Part (i) is a sine sum: the form sin Acos C+cos Asin C is the expansion of sin(A+C), and with A+C=90^∘ it must be 90^∘=1.
• Part (ii) is a cosine sum: likewise cos Acos C-sin Asin C is cos(A+C), which is 90^∘=0.
• Cross-check: these two patterns confirm the direct table substitution answers of one and zero, so you arrive at the same results by two separate routes without any extra work.
• Why it helps: although the angle-sum formulas are studied formally only later, spotting them here is a fast and reliable way to be sure that your standard-value substitution has been carried out correctly.

(i) sin(A+C)=90^∘=1;   (ii) cos(A+C)=90^∘=0.

Recognise the (5,12,13) triple at the end.

• Set up one unknown: let QR=x and PR=25-x; the x² terms cancel, leaving the single linear equation 625-50x=25, so x=12.
• Name the sides: the three sides are then PQ=5, QR=12 and PR=13, the classic (5,12,13) right triangle.
• Read from P: the ratios follow at once as sin P=1213, cos P=513 and tan P=125.
• Use it as a check: landing on a familiar triple is a reassuring sign; if the sides had not formed one, it would be worth rechecking the arithmetic before reading off the ratios.

sin P=1213, cos P=513, tan P=125 (a (5,12,13) triangle).

Justify with a one-line reason each.

• Counter-example for (i): a single case where the claim fails is enough, so quoting 60^∘=3 already disproves ``always less than one'' and you need write nothing more.
• Bound for (ii): the secant of an acute angle is always at least one, and the given value satisfies that bound, so it is achievable and the statement is true; cite the bound rather than just asserting the verdict.
• Notation for (iii) and (iv): point to the symbols plainly, since the cosine is written one way and the cosecant another, while the cotangent of an angle is a single symbol and not a product of two things.
• Bound for (v): the sine of any angle can never go above one, yet the stated value is larger than one, so no angle can satisfy it and the statement is false.
• The marking rule: every response should pair a verdict with a named value or a bound, because the examiner gives the marks for that justification and never for the bare verdict on its own.

Only (ii) is true; (i), (iii), (iv) and (v) are false, each with a one-line reason.

Rationalise carefully in parts (iii) and (iv).

• Substitute first: put in the standard table values for each part, after which the arithmetic is routine; the marks here are lost almost entirely on leftover surds sitting in the denominator, so plan to clear them.
• Part (iii): after substituting, the denominator is the sum 23+2, which combines over a common root, and multiplying both top and bottom by the conjugate removes every root and leaves the tidy value 32-68.
• Part (iv): reduce the messy fraction to the form (33-4)/(33+4) first, then multiply top and bottom by the conjugate of the denominator to land on 43-24311 with no surd left below the line.
• The easy parts: parts (i), (ii) and (v) need only the Pythagorean identity and some basic squaring, so they settle quickly into the values one, two and 6712 without any rationalising at all.
• Marker of a final answer: a denominator free of roots is the sign of a fully simplified answer, so always spend the extra line, since a form such as one over 2+23 would not be accepted as finished work in the board exam.

Values: 1, 2, 32-68, 43-24311, 6712.

Justify, do not just tick.

• Part (i): substituting the table values simplifies the expression to 32, which is recognised as 60^∘, so the correct option is the first one.
• Part (ii): the numerator becomes zero, giving the value zero; note that the secant-style distractor here is undefined, which is exactly the trap a hurried student falls into.
• Part (iii): only the angle zero makes the relation 2A=2sin A true, because both sides are then zero, whereas the other choices fail when you test them.
• Part (iv): the same substitution simplifies to 3, which is 60^∘, so that option is correct.
• The justification marks: a single line of working beside each option is enough, and it also guards against ticking a tempting choice whose value is actually undefined.

(i) A, (ii) D, (iii) A, (iv) C, each shown by substitution.

Verify both conditions hold.

• Form the system: matching each tangent to its standard angle turns the two equations into the pair A+B=60^∘ and A-B=30^∘.
• Solve the pair: adding and subtracting these gives the single solution A=45^∘ and B=15^∘.
• Check the limits: the sum stays within the allowed ninety degrees and the larger angle is indeed A, so both stated conditions are met and the answer is valid.
• Why it matters: the given inequalities are there to pin down a single solution, so quoting them in your check shows the examiner you used every piece of information.

A=45^∘, B=15^∘, satisfying A+B90^∘ and A>B.

Use one clear counter-example per false claim.

• The false sums: statement one fails at the angles thirty and sixty, since the sine of their sum is one but the separate sines add to more than one; the equal-ratio claim fails everywhere except at the single angle of forty-five degrees.
• The trend claims: the sine statement holds because sine rises steadily across the whole range from zero to ninety, while the cosine statement fails because cosine falls across that same range rather than rising.
• The undefined case: the cotangent at zero is a cosine over a sine, and the sine there is zero, so the ratio is a division by zero and is correctly undefined.
• The marking rule: pair every verdict with a value or a stated trend, since a single counter-example for a false claim, or the steady behaviour for a true one, is exactly what carries the justification marks.

(i) F, (ii) T, (iii) F, (iv) F, (v) T, each justified by a value or trend.

Take the positive root for acute angles.

• Sine: start from the cosecant identity and invert it to get the sine as one over the root of one plus the cotangent squared.
• Secant: write the tangent as the reciprocal of the cotangent, feed it into the secant identity, and the answer comes out over the cotangent.
• Tangent: this is the simplest of the three, being just the plain reciprocal of the cotangent with no root at all.
• Sign rule: because the angle is acute every ratio is positive in the first quadrant, so each square root carries a plus sign and no ambiguity remains; state that assumption to keep the answer rigorous.

sin A=1√1+cot² A, sec A=√1+cot² Acot A, tan A=1cot A.

Chain the identities in order.

• Anchor on cosine: the cosine is simply the reciprocal of the given secant, and this single value is what every later step leans on.
• Then the sine: feed that cosine into the Pythagorean identity to write the sine as a root over the secant.
• Then the rest: the tangent is sine over cosine, after which the cotangent and cosecant are just the reciprocals of the ratios already found.
• Why the order: following one fixed chain is far more reliable than jumping between identities, and it makes the common root factor appear naturally in three of the answers instead of being forced.

All five ratios expressed over sec A and √sec² A-1 as above.

Convert to sine and cosine when stuck.

• Part (i): this is just the secant-tangent identity scaled by nine, so the value is nine straight away with no real work.
• Part (ii): rewrite the brackets over sine and cosine, use the expansion of the squared sum, and the product collapses neatly to the value two.
• Parts (iii) and (iv): the third uses one minus the sine squared to land on the cosine, and the fourth is the ratio of the secant squared to the cosecant squared, which is the tangent squared.
• The general tactic: reach for the standard identities first because they are fast, but fall back to the slower sine-cosine route whenever a product looks tangled, since that route is longer yet never fails.

(i) B, (ii) C, (iii) D, (iv) D, each justified by an identity.

Match the method to the shape of the expression.

• Pick the move first: these ten proofs use a small set of standard moves, and choosing the right one quickly is the real skill being tested, so glance at each identity before touching the algebra and decide which single technique will carry it rather than wandering down a longer path.
• Difference of squares: this clears the first, fourth and sixth proofs, where you replace a sine squared or a cosine squared by one minus the other and then cancel the common factor; in the sixth the same idea works inside a square root once you multiply by the conjugate to remove the root cleanly.
• Common denominator: this settles the second and third proofs, where the second has its numerator collapse to twice one plus the sine, and the third uses the difference-of-cubes factor to cancel the awkward sine-minus-cosine term left in the denominator.
• Expand and group: this handles the eighth and ninth, where the eighth regroups its four squared terms into one plus four plus a cosecant squared plus a secant squared, while in the ninth each bracket becomes a single squared ratio over one side.
• Spot equal brackets: the seventh and tenth turn on the numerator and denominator brackets being equal up to a sign, so in the seventh one minus twice the sine squared matches twice the cosine squared minus one, and in the tenth the middle expression becomes minus the tangent, whose square is still the tangent squared.
• Keep it honest: write the left side on the left and work only on that side until it matches the right, never cross-multiplying across the equals sign, since that habit keeps every proof clean and honest, and it makes it easy for the examiner to follow your reasoning and award each of the step marks on offer.

All ten identities proved using sin²+cos²=1 and its two companions, one fitted technique per part.