NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.1 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.