Introduction to Trigonometry

Find the hypotenuse AC first, using the Pythagoras Theorem:

(i) Angle A: opposite side is BC, adjacent side is AB.

(ii) Angle C: opposite side is AB, adjacent side is BC (the sides swap roles for the OTHER acute angle).

Reading Fig. 8.13: \(\triangle\)PQR is right-angled at Q, with PQ = 12 cm and the hypotenuse PR = 13 cm. NCERT deliberately does not label QR — it must be found first, using the Pythagoras Theorem:

Now use the definitions. For \(\angle\)P, the opposite side is QR and the adjacent side is PQ; for \(\angle\)R, the opposite side is PQ and the adjacent side is QR:

So:

tan P − cot R = 0.

Use the fundamental identity \(\sin^2A+\cos^2A=1\) to find cos A first:

(taking the positive root, since A is acute and cos A must be positive.) Then:

From \(15\cot A=8\), \(\cot A=\dfrac{8}{15}=\dfrac{\text{side adjacent to A}}{\text{side opposite to A}}\). So in a right triangle with the angle A, take the side adjacent to A as 8k and the side opposite to A as 15k. By the Pythagoras Theorem, the hypotenuse is:

So:

\(\sec\theta=\dfrac{\text{hyp}}{\text{side adjacent to }\theta}=\dfrac{13}{12}\), so take the hypotenuse as 13k and the adjacent side as 12k. The opposite side follows from the Pythagoras Theorem:

Now every ratio follows from its definition:

The argument is the same as Example 2 (there it was sin B = sin Q; here it is cos A = cos B), run with the adjacent side instead of the opposite side.

Consider two right triangles ACB and BDA (or, more simply, drop a perpendicular from A and a perpendicular from B in a single reference triangle) — the standard way NCERT phrases it is via a single right \(\triangle\)ABC, right-angled at C, together with a second right \(\triangle\)PQR, right-angled at R, with cos A = cos P:

Since \(\cos A=\cos P\):

Using the Pythagoras Theorem to bring in the opposite sides:

Dividing these and combining with the ratio above (exactly as in Example 2) gives \(\dfrac{AC}{PR}=\dfrac{AB}{PQ}=\dfrac{BC}{QR}\), so by the SSS similarity criterion \(\triangle\)ABC \(\sim\) \(\triangle\)PQR, and therefore corresponding angles are equal: \(\angle\)A = \(\angle\)P (i.e. \(\angle\)A = \(\angle\)B when the second triangle's angle is relabelled B, as the question names it).

\(\cot\theta=\dfrac{7}{8}=\dfrac{\text{adjacent}}{\text{opposite}}\), so take the adjacent side as 7k and the opposite side as 8k. The hypotenuse:

So \(\sin\theta=\dfrac{8}{\sqrt{113}}\) and \(\cos\theta=\dfrac{7}{\sqrt{113}}\).

(i) Both brackets are a difference of squares: \((1+\sin\theta)(1-\sin\theta)=1-\sin^2\theta=\cos^2\theta\), and likewise \((1+\cos\theta)(1-\cos\theta)=1-\cos^2\theta=\sin^2\theta\). So the whole expression simplifies to \(\cot^2\theta\) BEFORE any number is even substituted:

(ii) This is exactly the same value, since part (i) simplified to \(\cot^2\theta\) identically:

From \(3\cot A=4\), \(\cot A=\dfrac43\), so \(\tan A=\dfrac34\). Taking the opposite side as 3k, adjacent side as 4k, the hypotenuse is:

So \(\sin A=\dfrac35\) and \(\cos A=\dfrac45\) (a 3-4-5 right triangle).

LHS:

RHS:

Since LHS = RHS = \(\dfrac{7}{25}\), yes, the identity \(\dfrac{1-\tan^2A}{1+\tan^2A}=\cos^2A-\sin^2A\) holds for this A (in fact it is a general identity, true for every acute A — dividing numerator and denominator of the LHS by \(\sec^2A\) turns it directly into the RHS).

\(\tan A=\dfrac{1}{\sqrt3}\) is the value of \(\tan 30^{\circ}\) from Table 8.1, so \(\angle A=30^{\circ}\). Since \(\angle B=90^{\circ}\), the angle sum gives \(\angle C=180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}\).

Reading straight from Table 8.1: \(\sin A=\dfrac12,\ \cos A=\dfrac{\sqrt3}{2},\ \sin C=\dfrac{\sqrt3}{2},\ \cos C=\dfrac12\).

(i)

(ii)

Note: both results are exactly what is expected, since A + C = 90° here: \(\sin A\cos C+\cos A\sin C\) is the expansion of \(\sin(A+C)=\sin90^{\circ}=1\), and \(\cos A\cos C-\sin A\sin C\) is the expansion of \(\cos(A+C)=\cos90^{\circ}=0\).

Let \(QR=x\), so \(PR=25-x\) (from PR + QR = 25). By the Pythagoras Theorem, \(PR^2=PQ^2+QR^2\):

The \(x^2\) terms cancel:

So \(QR=12\) cm and \(PR=25-12=13\) cm.

Now, for \(\angle\)P: the opposite side is QR, the adjacent side is PQ, and the hypotenuse is PR.

(i) False. tan A = opposite/adjacent, and there is no restriction forcing the opposite side to be shorter than the adjacent side. A direct counterexample:

(ii) True. sec A = hypotenuse/adjacent, so a value of \(\dfrac{12}{5}\) (which is \(\geq1\), as sec A must always be) corresponds to a genuine right triangle with hypotenuse 12k and adjacent side 5k:

a real, positive opposite side, so such a triangle (and such an angle A) genuinely exists.

(iii) False. “cos A” is the abbreviation for the cosine of angle A, not the cosecant. The cosecant of A is abbreviated “cosec A” (or “csc A”).

(iv) False. “cot A” is a single, indivisible symbol read as “the cotangent of angle A” — it is not the product of a quantity called “cot” with A, exactly as the Remark after the definitions on p.116 warns for sin A and cos A (“sin” separated from A has no meaning by itself).

(v) False. In a right triangle the hypotenuse is always the longest side, so the opposite side can never exceed the hypotenuse. This forces \(\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}\leq1\) for every angle \(\theta\), and \(\dfrac43>1\).

Every part is read straight off Table 8.1: \(\sin30^{\circ}=\cos60^{\circ}=\dfrac12\), \(\sin60^{\circ}=\cos30^{\circ}=\dfrac{\sqrt3}{2}\), \(\sin45^{\circ}=\cos45^{\circ}=\dfrac{1}{\sqrt2}\), \(\tan45^{\circ}=1\), \(\sec30^{\circ}=\dfrac{2}{\sqrt3}\), \(\text{cosec}\,60^{\circ}=\dfrac{2}{\sqrt3}\), \(\cot45^{\circ}=1\).

(i)

(ii)

(iii)

(iv)

Multiplying every term of the last fraction, top and bottom, by \(2\sqrt3\):

(v) The identity \(\sin^230^{\circ}+\cos^230^{\circ}=1\) makes the denominator trivial, so only the numerator needs evaluating:

(i) Substitute \(\tan30^{\circ}=\dfrac{1}{\sqrt3}\):

Comparing with Table 8.1, this equals \(\sin60^{\circ}=\dfrac{\sqrt3}{2}\). Correct option: (A) sin 60°.

(ii) Substitute \(\tan45^{\circ}=1\):

Correct option: (D) 0.

(iii) Test each option by substitution — only A = 0° can work in general, since \(\sin2A=2\sin A\cos A\), and this equals \(2\sin A\) only when \(\cos A=1\) (or \(\sin A=0\)), i.e. A = 0°:

The other options fail, e.g. at A = 30°: \(\sin60^{\circ}=\dfrac{\sqrt3}{2}\approx0.866\) but \(2\sin30^{\circ}=1\) — not equal. Correct option: (A) 0°.

(iv) Substitute \(\tan30^{\circ}=\dfrac{1}{\sqrt3}\):

Comparing with Table 8.1, this equals \(\tan60^{\circ}=\sqrt3\). Correct option: (C) tan 60°.

From Table 8.1, \(\tan60^{\circ}=\sqrt3\), and \(0^{\circ}\lt A+B\leq90^{\circ}\) is given directly, so:

Also from Table 8.1, \(\tan30^{\circ}=\dfrac{1}{\sqrt3}\), and \(A-B\) is acute (since \(A>B\) and both are angles of this kind):

Add and subtract the two equations, exactly as in Example 8:

So A = 45° and B = 15°.

(i) False. A single counterexample is enough: take A = B = 30°.

Since \(\dfrac{\sqrt3}{2}\approx0.866\neq1\), sin (A + B) ≠ sin A + sin B here, so the claimed identity is false in general.

(ii) True. As \(\theta\) runs from 0° to 90°, the values in Table 8.1 climb steadily: \(0,\ \tfrac12,\ \tfrac{1}{\sqrt2},\ \tfrac{\sqrt3}{2},\ 1\) — each value is larger than the one before it, matching the Remark stated right under Table 8.1.

(iii) False. cos \(\theta\) does the OPPOSITE — it steadily decreases as \(\theta\) increases from 0° to 90°: \(1,\ \tfrac{\sqrt3}{2},\ \tfrac{1}{\sqrt2},\ \tfrac12,\ 0\) (same Remark, second half).

(iv) False. They are equal only at the single angle \(\theta=45^{\circ}\) (where both equal \(\tfrac{1}{\sqrt2}\)); at any other angle they differ — e.g. at \(\theta=30^{\circ}\), \(\sin30^{\circ}=\tfrac12\) but \(\cos30^{\circ}=\tfrac{\sqrt3}{2}\).

(v) True. \(\cot A=\dfrac{\cos A}{\sin A}\), and \(\sin0^{\circ}=0\), so this is division by zero — exactly why Table 8.1 marks cot 0° as “Not defined”.

Start from the identity \(\text{cosec}^2A=1+\cot^2A\) to bring in sin A:

Taking the positive root (A acute):

Then sin A follows since \(\text{cosec}\,A=\dfrac{1}{\sin A}\):

tan A is immediate, since cot A and tan A are reciprocals:

For sec A, use \(\sec^2A=1+\tan^2A\) with the tan A just found:

cos A is immediate, since sec A and cos A are reciprocals:

For sin A, start from \(\sin^2A+\cos^2A=1\):

tan A follows from \(\sec^2A=1+\tan^2A\):

Finally cosec A and cot A are the reciprocals of sin A and tan A just found:

(i) Factor out the 9 and use \(\sec^2A-\tan^2A=1\) (from \(\sec^2A=1+\tan^2A\)):

Correct option: (B) 9.

(ii) Write everything over sin \(\theta\) and cos \(\theta\):

The numerator is a difference of squares, \((\sin\theta+\cos\theta)^2-1^2\), and \(1=\sin^2\theta+\cos^2\theta\), so \((\sin\theta+\cos\theta)^2-1=2\sin\theta\cos\theta\):

Correct option: (C) 2.

(iii) Write \(\sec A+\tan A=\dfrac{1+\sin A}{\cos A}\), so the product becomes a difference of squares:

Correct option: (D) cos A.

(iv) Write \(1+\tan^2A=\sec^2A\) and \(1+\cot^2A=\text{cosec}^2A\):

Correct option: (D) \(\tan^2A\).

(i) Write both ratios over sin \(\theta\):

Multiply top and bottom by \((1+\cos\theta)\), using \(1-\cos^2\theta=\sin^2\theta\):

(ii) Combine the two fractions over the common denominator \(\cos A(1+\sin A)\):

(iii) Following the hint, write \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) and \(\cot\theta=\dfrac{\cos\theta}{\sin\theta}\):

Flip the sign of the bottom bracket in the second fraction so both denominators read \((\sin\theta-\cos\theta)\):

Add the two fractions (common denominator \(\sin\theta\cos\theta(\sin\theta-\cos\theta)\)):

Since \(\cos^3\theta-\sin^3\theta=(\cos\theta-\sin\theta)(\cos^2\theta+\sin\theta\cos\theta+\sin^2\theta)=(\cos\theta-\sin\theta)(1+\sin\theta\cos\theta)\), and \((\cos\theta-\sin\theta)=-(\sin\theta-\cos\theta)\):

(iv) Simplify each side on its own, as the hint suggests.

LHS:

RHS: write \(\sin^2A=1-\cos^2A\) as a difference of squares:

LHS = RHS = \(1+\cos A\).

(v) Divide every term, top and bottom, by sin A, so cos A / sin A becomes cot A and 1 / sin A becomes cosec A:

Group the numerator around \((\cot A+\text{cosec}\,A)\) and the denominator around \((\cot A-\text{cosec}\,A)\) — the two group differently because of the sign on cosec A:

Multiply the numerator and the denominator separately by that same bracket, \((\cot A+\text{cosec}\,A)\), so the identity \(\text{cosec}^2A-\cot^2A=1\) (from \(\text{cosec}^2A=1+\cot^2A\)) can be used on the denominator next:

Expand each:

Since \(\cot^2A-\text{cosec}^2A=-1\), the denominator is \((\cot A+\text{cosec}\,A)-1\) — the SAME bracket both numerator terms share, so it cancels. Writing \(K=\cot A+\text{cosec}\,A\) just for this step, to keep the fraction readable:

(vi) Multiply inside the square root, top and bottom, by \((1+\sin A)\):

(vii) Factor \(\sin\theta\) from the numerator and \(\cos\theta\) from the denominator:

Since \(1-2\sin^2\theta=\cos^2\theta-\sin^2\theta\) and \(2\cos^2\theta-1=\cos^2\theta-\sin^2\theta\) (both equal, using \(\sin^2\theta+\cos^2\theta=1\)), the two brackets are equal and cancel:

(viii) Expand both squares first:

Group the sin²+cos² pair and the cosec²+sec² pair:

Since \(\text{cosec}^2A=1+\cot^2A\) and \(\sec^2A=1+\tan^2A\):

(ix) Simplify each side on its own, as the hint suggests.

LHS:

RHS:

LHS = RHS = \(\sin A\cos A\).

(x) First show the outer equality, \(\dfrac{1+\tan^2A}{1+\cot^2A}=\tan^2A\), exactly as in Exercise 8.3 Q3(iv) above:

Now the middle expression. Write \(\cot A=\dfrac{1}{\tan A}\) throughout:

Multiplying top and bottom of the inner fraction by tan A:

So all three expressions equal \(\tan^2A\), as required.