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Mathematics · Introduction to Trigonometry

NCERT Examples

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

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.

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.

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.

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).

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.

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.