Mathematics · Introduction to Trigonometry

NCERT Examples

All 12 worked examples from Introduction to Trigonometry, with independently verified solutions and direct navigation.

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Mathematics · Chapter 8 · NCERT Examples

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NCERT Example

Example 1

trigonometric ratios · Pythagoras theorem

Given tan A = 4/3, find the other trigonometric ratios of angle A.

Right triangle ABC, right-angled at B, with AB equal to 3k and BC equal to 4k.
Figure 8-8 from the supplied NCERT chapter.

Solution

Board-exam working:

tan⁡A=BCAB=43\tan A=\frac{BC}{AB}=\frac43

Let BC = 4k and AB = 3k.

AC=AB2+BC2=(3k)2+(4k)2=5kAC=\sqrt{AB^2+BC^2}=\sqrt{(3k)^2+(4k)^2}=5k
sin⁡A=BCAC=45\sin A=\frac{BC}{AC}=\frac45
cos⁡A=ABAC=35\cos A=\frac{AB}{AC}=\frac35
cot⁡A=34\cot A=\frac34
cosec⁡A=54\cosec A=\frac54
sec⁡A=53\sec A=\frac53

Final answer

sin⁡A=45, cos⁡A=35, cot⁡A=34, cosec⁡A=54, sec⁡A=53\sin A=\frac45,\ \cos A=\frac35,\ \cot A=\frac34,\ \cosec A=\frac54,\ \sec A=\frac53

NCERT Example

Example 2

equal sine ratios · similar triangles

If B and Q are acute angles such that sin B = sin Q, prove that B = Q.

Two right triangles ABC and PQR with equal acute angles B and Q marked for comparison.
Figure 8-9 from the supplied NCERT chapter.

Solution

Board-exam working:

Consider right triangles ABC and PQR as shown.

sin⁡B=ACAB,sin⁡Q=PRPQ\sin B=\frac{AC}{AB},\qquad \sin Q=\frac{PR}{PQ}
ACAB=PRPQ\frac{AC}{AB}=\frac{PR}{PQ}
ACPR=ABPQ=k\frac{AC}{PR}=\frac{AB}{PQ}=k

By the Pythagoras theorem:

BC=AB2−AC2,QR=PQ2−PR2BC=\sqrt{AB^2-AC^2},\qquad QR=\sqrt{PQ^2-PR^2}
BCQR=k2PQ2−k2PR2PQ2−PR2=k\frac{BC}{QR}=\frac{\sqrt{k^2PQ^2-k^2PR^2}}{\sqrt{PQ^2-PR^2}}=k
ACPR=ABPQ=BCQR\frac{AC}{PR}=\frac{AB}{PQ}=\frac{BC}{QR}

Therefore, △ACB ∼ △PRQ by SSS similarity. Corresponding angles are equal.

∠B=∠Q\angle B=\angle Q

Final answer

Hence, ∠B = ∠Q.

NCERT Example

Example 3

trigonometric ratios · squares

In right triangle ACB, AB = 29 units, BC = 21 units and angle ABC = θ. Find (i) cos²θ + sin²θ and (ii) cos²θ − sin²θ.

Right triangle ACB, right-angled at C, with BC 21 units, AB 29 units, and angle theta at B.
Figure 8-10 from the supplied NCERT chapter.

Solution

Board-exam working:

AC=AB2−BC2=292−212=20AC=\sqrt{AB^2-BC^2}=\sqrt{29^2-21^2}=20
sin⁡θ=2029,cos⁡θ=2129\sin\theta=\frac{20}{29},\qquad \cos\theta=\frac{21}{29}

(i)

cos⁡2θ+sin⁡2θ=(2129)2+(2029)2=841841=1\cos^2\theta+\sin^2\theta=\left(\frac{21}{29}\right)^2+\left(\frac{20}{29}\right)^2=\frac{841}{841}=1

(ii)

cos⁡2θ−sin⁡2θ=212−202292=41841\cos^2\theta-\sin^2\theta=\frac{21^2-20^2}{29^2}=\frac{41}{841}

Final answer

(i) 1

(ii) 41841\text{(ii) }\frac{41}{841}

NCERT Example

Example 4

tan A · verification

In right triangle ABC, right-angled at B, tan A = 1. Verify that 2 sin A cos A = 1.

Isosceles right triangle ABC, right-angled at B, used to verify a trigonometric relation.
Figure 8-11 from the supplied NCERT chapter.

Solution

Board-exam working:

tan⁡A=BCAB=1⇒BC=AB\tan A=\frac{BC}{AB}=1\Rightarrow BC=AB

Let AB = BC = k.

AC=k2+k2=k2AC=\sqrt{k^2+k^2}=k\sqrt2
sin⁡A=kk2=12\sin A=\frac{k}{k\sqrt2}=\frac1{\sqrt2}
cos⁡A=kk2=12\cos A=\frac{k}{k\sqrt2}=\frac1{\sqrt2}
2sin⁡Acos⁡A=2(12)(12)=12\sin A\cos A=2\left(\frac1{\sqrt2}\right)\left(\frac1{\sqrt2}\right)=1

Final answer

Verified: 2 sin A cos A = 1.

NCERT Example

Example 5

right triangle · trigonometric ratios

In right triangle OPQ, right-angled at P, OP = 7 cm and OQ − PQ = 1 cm. Find sin Q and cos Q.

Right triangle OPQ, right-angled at P, with OP equal to 7 centimetres.
Figure 8-12 from the supplied NCERT chapter.

Solution

Board-exam working:

Let PQ = x cm. Then OQ = x + 1 cm.

OQ2=OP2+PQ2OQ^2=OP^2+PQ^2
(x+1)2=72+x2(x+1)^2=7^2+x^2
x2+2x+1=49+x2x^2+2x+1=49+x^2
2x=48⇒x=242x=48\Rightarrow x=24
PQ=24 cm,OQ=25 cmPQ=24\text{ cm},\qquad OQ=25\text{ cm}
sin⁡Q=OPOQ=725\sin Q=\frac{OP}{OQ}=\frac7{25}
cos⁡Q=PQOQ=2425\cos Q=\frac{PQ}{OQ}=\frac{24}{25}

Final answer

sin⁡Q=725,cos⁡Q=2425\sin Q=\frac7{25},\qquad \cos Q=\frac{24}{25}

NCERT Example

Example 6

special angles · 30 degrees

In right triangle ABC, right-angled at B, AB = 5 cm and angle ACB = 30°. Find BC and AC.

Right triangle ABC, right-angled at B, with AB 5 centimetres and angle C equal to 30 degrees.
Figure 8-19 from the supplied NCERT chapter.

Solution

Board-exam working:

tan⁡30∘=ABBC\tan30^\circ=\frac{AB}{BC}
13=5BC\frac1{\sqrt3}=\frac5{BC}
BC=53 cmBC=5\sqrt3\text{ cm}
sin⁡30∘=ABAC\sin30^\circ=\frac{AB}{AC}
12=5AC\frac12=\frac5{AC}
AC=10 cmAC=10\text{ cm}

Final answer

BC = 5√3 cm and AC = 10 cm.

NCERT Example

Example 7

special angles · complementary angles

In right triangle PQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Find angles QPR and PRQ.

Right triangle PQR, right-angled at Q, with PQ 3 centimetres and PR 6 centimetres.
Figure 8-20 from the supplied NCERT chapter.

Solution

Board-exam working:

sin⁡R=PQPR=36=12\sin R=\frac{PQ}{PR}=\frac36=\frac12
∠PRQ=30∘\angle PRQ=30^\circ

The two acute angles of a right triangle are complementary.

∠QPR=90∘−30∘=60∘\angle QPR=90^\circ-30^\circ=60^\circ

Final answer

∠QPR = 60° and ∠PRQ = 30°.

NCERT Example

Example 8

special angles · simultaneous equations

If sin(A − B) = 1/2 and cos(A + B) = 1/2, where 0° < A + B ≤ 90° and A > B, find A and B.

Special angles in A − B and A + BOne right triangle illustrates A minus B equal to 30 degrees and A plus B equal to 60 degrees. These are angle expressions, not vertex names.√312A − B = 30°A + B = 60°
Special angles in A − B and A + B — explanatory sketch (not to scale).

Solution

Board-exam working:

sin⁡(A−B)=12⇒A−B=30∘\sin(A-B)=\frac12\Rightarrow A-B=30^\circ
cos⁡(A+B)=12⇒A+B=60∘\cos(A+B)=\frac12\Rightarrow A+B=60^\circ

Adding the equations:

2A=90∘⇒A=45∘2A=90^\circ\Rightarrow A=45^\circ

Substituting in A + B = 60°:

45∘+B=60∘⇒B=15∘45^\circ+B=60^\circ\Rightarrow B=15^\circ

Final answer

A = 45° and B = 15°.

NCERT Example

Example 9

trigonometric identities · in terms of sin

Express cos A, tan A and sec A in terms of sin A.

Expressing ratios through sin AA right triangle marks angle A, its opposite side, adjacent side and hypotenuse so the trigonometric ratios can be read directly.adjacentoppositehypotenuseAABC
Expressing ratios through sin A — explanatory sketch (not to scale).

Solution

Board-exam working:

sin⁡2A+cos⁡2A=1\sin^2A+\cos^2A=1
cos⁡2A=1−sin⁡2A\cos^2A=1-\sin^2A

Since A is acute, cos A is positive.

cos⁡A=1−sin⁡2A\cos A=\sqrt{1-\sin^2A}
tan⁡A=sin⁡Acos⁡A=sin⁡A1−sin⁡2A\tan A=\frac{\sin A}{\cos A}=\frac{\sin A}{\sqrt{1-\sin^2A}}
sec⁡A=1cos⁡A=11−sin⁡2A\sec A=\frac1{\cos A}=\frac1{\sqrt{1-\sin^2A}}

Final answer

cos⁡A=1−sin⁡2A,tan⁡A=sin⁡A1−sin⁡2A,sec⁡A=11−sin⁡2A\cos A=\sqrt{1-\sin^2A},\quad \tan A=\frac{\sin A}{\sqrt{1-\sin^2A}},\quad \sec A=\frac1{\sqrt{1-\sin^2A}}

NCERT Example

Example 10

trigonometric identity · proof

Prove that sec A(1 − sin A)(sec A + tan A) = 1.

Identity using sin A, tan A and sec AA right triangle marks angle A, its opposite side, adjacent side and hypotenuse so the trigonometric ratios can be read directly.adjacentoppositehypotenuseAABC
Identity using sin A, tan A and sec A — explanatory sketch (not to scale).

Solution

Board-exam working:

Start with the left-hand side:

LHS=sec⁡A(1−sin⁡A)(sec⁡A+tan⁡A)\mathrm{LHS}=\sec A(1-\sin A)(\sec A+\tan A)
=1cos⁡A(1−sin⁡A)(1cos⁡A+sin⁡Acos⁡A)=\frac1{\cos A}(1-\sin A)\left(\frac1{\cos A}+\frac{\sin A}{\cos A}\right)
=(1−sin⁡A)(1+sin⁡A)cos⁡2A=\frac{(1-\sin A)(1+\sin A)}{\cos^2A}
=1−sin⁡2Acos⁡2A=\frac{1-\sin^2A}{\cos^2A}
=cos⁡2Acos⁡2A=1=RHS=\frac{\cos^2A}{\cos^2A}=1=\mathrm{RHS}

Final answer

Hence proved.

NCERT Example

Example 11

trigonometric identity · proof

Prove that (cot A − cos A)/(cot A + cos A) = (cosec A − 1)/(cosec A + 1).

Identity using cot A and cosec AA right triangle marks angle A, its opposite side, adjacent side and hypotenuse so the trigonometric ratios can be read directly.adjacentoppositehypotenuseAABC
Identity using cot A and cosec A — explanatory sketch (not to scale).

Solution

Board-exam working:

LHS=cot⁡A−cos⁡Acot⁡A+cos⁡A\mathrm{LHS}=\frac{\cot A-\cos A}{\cot A+\cos A}
=cos⁡Asin⁡A−cos⁡Acos⁡Asin⁡A+cos⁡A=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}
=cos⁡A(1sin⁡A−1)cos⁡A(1sin⁡A+1)=\frac{\cos A\left(\frac1{\sin A}-1\right)}{\cos A\left(\frac1{\sin A}+1\right)}
=cosec⁡A−1cosec⁡A+1=RHS=\frac{\cosec A-1}{\cosec A+1}=\mathrm{RHS}

Final answer

Hence proved.

NCERT Example

Example 12

trigonometric identity · secant tangent

Prove that (sin θ − cos θ + 1)/(sin θ + cos θ − 1) = 1/(sec θ − tan θ), using sec²θ = 1 + tan²θ.

Identity using sec θ and tan θA right triangle marks angle θ, its opposite side, adjacent side and hypotenuse so the trigonometric ratios can be read directly.adjacentoppositehypotenuseθABC
Identity using sec θ and tan θ — explanatory sketch (not to scale).

Solution

Board-exam working:

Divide the numerator and denominator of the left-hand side by cos θ:

LHS=tan⁡θ−1+sec⁡θtan⁡θ+1−sec⁡θ\mathrm{LHS}=\frac{\tan\theta-1+\sec\theta}{\tan\theta+1-\sec\theta}
=(tan⁡θ+sec⁡θ)−1(tan⁡θ−sec⁡θ)+1=\frac{(\tan\theta+\sec\theta)-1}{(\tan\theta-\sec\theta)+1}

Multiply numerator and denominator by tan θ − sec θ:

={(tan⁡θ+sec⁡θ)−1}(tan⁡θ−sec⁡θ){(tan⁡θ−sec⁡θ)+1}(tan⁡θ−sec⁡θ)=\frac{\{(\tan\theta+\sec\theta)-1\}(\tan\theta-\sec\theta)}{\{(\tan\theta-\sec\theta)+1\}(\tan\theta-\sec\theta)}
=tan⁡2θ−sec⁡2θ−tan⁡θ+sec⁡θ(tan⁡θ−sec⁡θ+1)(tan⁡θ−sec⁡θ)=\frac{\tan^2\theta-\sec^2\theta-\tan\theta+\sec\theta}{(\tan\theta-\sec\theta+1)(\tan\theta-\sec\theta)}
=−1−tan⁡θ+sec⁡θ(tan⁡θ−sec⁡θ+1)(tan⁡θ−sec⁡θ)=\frac{-1-\tan\theta+\sec\theta}{(\tan\theta-\sec\theta+1)(\tan\theta-\sec\theta)}
=−1tan⁡θ−sec⁡θ=1sec⁡θ−tan⁡θ=RHS=\frac{-1}{\tan\theta-\sec\theta}=\frac1{\sec\theta-\tan\theta}=\mathrm{RHS}

Final answer

Hence proved.

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