Introduction to Trigonometry — CBSE Class 10 Maths Textbook Solutions

First find the hypotenuse $AC$AC using Pythagoras' theorem:
$AC=\sqrt{AB^2+BC^2}=\sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=25\text{ cm}$AC=√AB^2+BC^2=√24^2+7^2=√576+49=√625=25 cm

(i) With respect to $\angle A$A: opposite side $=BC=7$=BC=7, adjacent side $=AB=24$=AB=24, hypotenuse $=25$=25.
$\sin A=\frac{BC}{AC}=\frac{7}{25},\qquad \cos A=\frac{AB}{AC}=\frac{24}{25}$A=BC/AC=7/25, A=AB/AC=24/25

(ii) With respect to $\angle C$C: opposite side $=AB=24$=AB=24, adjacent side $=BC=7$=BC=7, hypotenuse $=25$=25.
$\sin C=\frac{AB}{AC}=\frac{24}{25},\qquad \cos C=\frac{BC}{AC}=\frac{7}{25}$C=AB/AC=24/25, C=BC/AC=7/25

Since $\triangle PQR$PQR is right-angled at $Q$Q, $\angle P+\angle R=90^\circ$P+ R=90^.

With respect to $\angle P$P: opposite $=QR=12$=QR=12, adjacent $=PQ=5$=PQ=5, so $\tan P=\dfrac{QR}{PQ}=\dfrac{12}{5}$P=QR/PQ=12/5.

With respect to $\angle R$R: opposite $=PQ=5$=PQ=5, adjacent $=QR=12$=QR=12, so $\cot R=\dfrac{\text{adjacent to }R}{\text{opposite to }R}=\dfrac{QR}{PQ}=\dfrac{12}{5}$R=adjacent to R/opposite to R=QR/PQ=12/5.

$\tan P-\cot R=\frac{12}{5}-\frac{12}{5}=0$P- R=12/5-12/5=0

(This is true in general for any right triangle right-angled at $Q$Q: since $\angle P$P and $\angle R$R are complementary, $\cot R=\cot(90^\circ-P)=\tan P$R=(90^-P)= P, so $\tan P-\cot R=0$P- R=0 always.)

Since $\sin A=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac34$A=opposite/hypotenuse=34, take opposite side $=3k$=3k, hypotenuse $=4k$=4k for some $k>0$k>0.

By Pythagoras, adjacent side $=\sqrt{(4k)^2-(3k)^2}=\sqrt{16k^2-9k^2}=\sqrt{7k^2}=\sqrt7\,k$=√(4k)^2-(3k)^2=√16k^2-9k^2=√7k^2=7\,k.

$\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt7\,k}{4k}=\frac{\sqrt7}{4}$A=adjacent/hypotenuse=7\,k/4k=7/4
$\tan A=\frac{\text{opposite}}{\text{adjacent}}=\frac{3k}{\sqrt7\,k}=\frac{3}{\sqrt7}$A=opposite/adjacent=3k/7\,k=3/7

$\cot A=\dfrac{8}{15}=\dfrac{\text{adjacent}}{\text{opposite}}$A=8/15=adjacent/opposite. Take adjacent $=8k$=8k, opposite $=15k$=15k.

By Pythagoras, hypotenuse $=\sqrt{(8k)^2+(15k)^2}=\sqrt{64k^2+225k^2}=\sqrt{289k^2}=17k$=√(8k)^2+(15k)^2=√64k^2+225k^2=√289k^2=17k.

$\sin A=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{15k}{17k}=\frac{15}{17}$A=opposite/hypotenuse=15k/17k=15/17
$\sec A=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{17k}{8k}=\frac{17}{8}$A=hypotenuse/adjacent=17k/8k=17/8

$\sec\theta=\dfrac{13}{12}=\dfrac{\text{hypotenuse}}{\text{adjacent}}$=13/12=hypotenuse/adjacent. Take hypotenuse $=13k$=13k, adjacent $=12k$=12k.

By Pythagoras, opposite $=\sqrt{(13k)^2-(12k)^2}=\sqrt{169k^2-144k^2}=\sqrt{25k^2}=5k$=√(13k)^2-(12k)^2=√169k^2-144k^2=√25k^2=5k.

$\sin\theta=\frac{5k}{13k}=\frac{5}{13},\quad \cos\theta=\frac{12k}{13k}=\frac{12}{13},\quad \tan\theta=\frac{5k}{12k}=\frac{5}{12}$=5k/13k=5/13, =12k/13k=12/13, =5k/12k=5/12
$\cot\theta=\frac{12}{5},\quad \csc\theta=\frac{13}{5},\quad \sec\theta=\frac{13}{12}\ \text{(given)}$=12/5, =13/5, =13/12 (given)

Since $\angle A$A is acute, $\cos A=\sqrt{1-\sin^2A}$A=√1-^2A is well defined with a positive value, and similarly for $\angle B$B.

Given $\cos A=\cos B$A= B. Using the identity $\sin^2\theta+\cos^2\theta=1$^2+^2=1:
$\sin A=\sqrt{1-\cos^2A}=\sqrt{1-\cos^2B}=\sin B$A=√1-^2A=√1-^2B= B
(taking the positive square root since both angles are acute, so both sines are positive).

So $\sin A=\sin B$A= B and $\cos A=\cos B$A= B simultaneously. On the interval $0^\circ$0^ to $90^\circ$90^, both $\sin\theta$ and $\cos\theta$ are strictly monotonic (one-to-one) functions, so equal cosines (or equal sines) for acute angles forces the angles themselves to be equal.

Hence $\angle A = \angle B$A = B. Proved.

(i) Using $(1+x)(1-x)=1-x^2$(1+x)(1-x)=1-x^2:
$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}=\frac{1-\sin^2\theta}{1-\cos^2\theta}=\frac{\cos^2\theta}{\sin^2\theta}=\cot^2\theta$(1+)(1-)/(1+)(1-)=1-^2/1-^2=^2/^2=^2
So both parts reduce to the same value, $\cot^2\theta=\left(\dfrac78\right)^2=\dfrac{49}{64}$^2=(78)^2=49/64.

(ii) $\cot^2\theta=\left(\dfrac78\right)^2=\dfrac{49}{64}$^2=(78)^2=49/64.

Both (i) and (ii) equal $\dfrac{49}{64}$49/64.

$\cot A=\dfrac43 \implies \tan A=\dfrac34$A=43 A=34. Take opposite $=3$=3, adjacent $=4$=4, hypotenuse $=\sqrt{3^2+4^2}=5$=√3^2+4^2=5.

So $\sin A=\dfrac35$A=35, $\cos A=\dfrac45$A=45.

LHS:
$\frac{1-\tan^2A}{1+\tan^2A}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{\frac{7}{16}}{\frac{25}{16}}=\frac{7}{25}$1-^2A/1+^2A=1-9/161+9/16=7/1625/16=7/25

RHS:
$\cos^2A-\sin^2A=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$^2A-^2A=16/25-9/25=7/25

Since LHS $=$= RHS $=\dfrac{7}{25}$=7/25, the identity holds (Yes).

Since the triangle is right-angled at $B$B, $\angle A+\angle C=90^\circ$A+ C=90^. Also $\tan A=\dfrac{1}{\sqrt3}$A=1/3 means $\angle A=30^\circ$A=30^, so $\angle C=60^\circ$C=60^.

$\sin A=\frac12,\ \cos A=\frac{\sqrt3}{2},\ \sin C=\frac{\sqrt3}{2},\ \cos C=\frac12$A=12, A=3/2, C=3/2, C=12

(i)
$\sin A\cos C+\cos A\sin C=\left(\frac12\right)\left(\frac12\right)+\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right)=\frac14+\frac34=1$A C+ A C=(12)(12)+(3/2)(3/2)=14+34=1

(ii)
$\cos A\cos C-\sin A\sin C=\left(\frac{\sqrt3}{2}\right)\left(\frac12\right)-\left(\frac12\right)\left(\frac{\sqrt3}{2}\right)=\frac{\sqrt3}{4}-\frac{\sqrt3}{4}=0$A C- A C=(3/2)(12)-(12)(3/2)=3/4-3/4=0

(These match the general identities $\sin A\cos C+\cos A\sin C=\sin(A+C)=\sin90^\circ=1$A C+ A C=(A+C)=90^=1 and $\cos A\cos C-\sin A\sin C=\cos(A+C)=\cos90^\circ=0$A C- A C=(A+C)=90^=0, which hold for any complementary pair $A, C$A, C.)

Let $QR=x$QR=x, so $PR=25-x$PR=25-x. By Pythagoras' theorem in $\triangle PQR$PQR:
$PQ^2+QR^2=PR^2$PQ^2+QR^2=PR^2
$5^2+x^2=(25-x)^2$5^2+x^2=(25-x)^2
$25+x^2=625-50x+x^2$25+x^2=625-50x+x^2
$25=625-50x$25=625-50x
$50x=600 \implies x=12$50x=600 x=12

So $QR=12$QR=12 cm and $PR=25-12=13$PR=25-12=13 cm.

With respect to $\angle P$P: opposite $=QR=12$=QR=12, adjacent $=PQ=5$=PQ=5, hypotenuse $=PR=13$=PR=13.
$\sin P=\frac{12}{13},\qquad \cos P=\frac{5}{13},\qquad \tan P=\frac{12}{5}$P=12/13, P=5/13, P=12/5

(i) False. $\tan A$A can exceed $1$1: e.g. $\tan60^\circ=\sqrt3>1$60^=3>1. It equals $1$1 at $45^\circ$45^ and is less than $1$1 only for $A<45^\circ$A<45^.

(ii) True. $\sec A=\dfrac{\text{hypotenuse}}{\text{adjacent}}\ge 1$A=hypotenuse/adjacent≥ 1 always (hypotenuse is the longest side), and $\dfrac{12}{5}=2.4\ge1$12/5=2.4≥1, so it is achievable — take a right triangle with adjacent $=5$=5, hypotenuse $=12$=12 (opposite $=\sqrt{144-25}=\sqrt{119}$=√144-25=√119).

(iii) False. $\cos A$A is the abbreviation for the cosine of $A$A, not the cosecant; the cosecant of $A$A is abbreviated $\text{cosec } A$cosec A (or $\csc A$A).

(iv) False. $\cot A$A is a single symbol denoting the cotangent of angle $A$A; it is not the product of $\cot$ and $A$A (just as $\sin A$A is not "$\sin$ times $A$A").

(v) False. Since the hypotenuse is always the longest side of a right triangle, $\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}\le1$=opposite/hypotenuse≤1 for every angle $\theta$. As $\dfrac43>1$43>1, no angle can give $\sin\theta=\dfrac43$=43.

Use the standard values: $\sin30^\circ=\frac12,\cos30^\circ=\frac{\sqrt3}{2},\tan30^\circ=\frac{1}{\sqrt3};\ \sin45^\circ=\cos45^\circ=\frac{1}{\sqrt2},\tan45^\circ=1;\ \sin60^\circ=\frac{\sqrt3}{2},\cos60^\circ=\frac12,\tan60^\circ=\sqrt3;\ \sec30^\circ=\frac{2}{\sqrt3},\csc30^\circ=2,\csc60^\circ=\frac{2}{\sqrt3}$30^=12,30^=3/2,30^=1/3; 45^=45^=1/2,45^=1; 60^=3/2,60^=12,60^=3; 30^=2/3,30^=2,60^=2/3.

(i) $\sin60^\circ\cos30^\circ+\sin30^\circ\cos60^\circ=\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac12\cdot\dfrac12=\dfrac34+\dfrac14=1$60^30^+30^60^=3/2·3/2+12·12=34+14=1

(ii) $2\tan^245^\circ+\cos^230^\circ-\sin^260^\circ=2(1)^2+\left(\dfrac{\sqrt3}{2}\right)^2-\left(\dfrac{\sqrt3}{2}\right)^2=2+\dfrac34-\dfrac34=2$2^245^+^230^-^260^=2(1)^2+(3/2)^2-(3/2)^2=2+34-34=2

(iii) $\dfrac{\cos45^\circ}{\sec30^\circ+\csc30^\circ}=\dfrac{1/\sqrt2}{\frac{2}{\sqrt3}+2}=\dfrac{1/\sqrt2}{\frac{2+2\sqrt3}{\sqrt3}}=\dfrac{\sqrt3}{\sqrt2(2+2\sqrt3)}=\dfrac{\sqrt3}{2\sqrt2(1+\sqrt3)}$45^/30^+30^=1/22/3+2=1/22+23/3=3/2(2+23)=3/22(1+3). Rationalising by $(\sqrt3-1)$(3-1): $=\dfrac{\sqrt3(\sqrt3-1)}{2\sqrt2(3-1)}=\dfrac{3-\sqrt3}{4\sqrt2}=\dfrac{3\sqrt2-\sqrt6}{8}$=3(3-1)/22(3-1)=3-3/42=32-6/8

(iv) Numerator $=\dfrac12+1-\dfrac{2}{\sqrt3}=\dfrac32-\dfrac{2}{\sqrt3}=\dfrac{3\sqrt3-4}{2\sqrt3}$=12+1-2/3=32-2/3=33-4/23. Denominator $=\dfrac{2}{\sqrt3}+\dfrac12+1=\dfrac{4+3\sqrt3}{2\sqrt3}$=2/3+12+1=4+33/23. So the ratio $=\dfrac{3\sqrt3-4}{3\sqrt3+4}$=33-4/33+4. Multiplying numerator and denominator by $(3\sqrt3-4)$(33-4): $=\dfrac{(3\sqrt3-4)^2}{(3\sqrt3)^2-4^2}=\dfrac{27-24\sqrt3+16}{27-16}=\dfrac{43-24\sqrt3}{11}$=(33-4)^2/(33)^2-4^2=27-243+16/27-16=43-243/11

(v) Numerator $=5\left(\dfrac12\right)^2+4\left(\dfrac{2}{\sqrt3}\right)^2-1^2=5\cdot\dfrac14+4\cdot\dfrac43-1=\dfrac54+\dfrac{16}{3}-1$=5(12)^2+4(2/3)^2-1^2=5·14+4·43-1=54+16/3-1. Common denominator $12$12: $\dfrac{15}{12}+\dfrac{64}{12}-\dfrac{12}{12}=\dfrac{67}{12}$15/12+64/12-12/12=67/12. Denominator $=\sin^230^\circ+\cos^230^\circ=1$=^230^+^230^=1 (identity). So the value is $\dfrac{67}{12}$67/12.

Answers: (i) $1$1 (ii) $2$2 (iii) $\dfrac{3\sqrt2-\sqrt6}{8}$32-6/8 (iv) $\dfrac{43-24\sqrt3}{11}$43-243/11 (v) $\dfrac{67}{12}$67/12

(i) $\tan30^\circ=\dfrac{1}{\sqrt3}$30^=1/3, so $\tan^230^\circ=\dfrac13$^230^=13.
$\frac{2\tan30^\circ}{1+\tan^230^\circ}=\frac{2/\sqrt3}{1+1/3}=\frac{2/\sqrt3}{4/3}=\frac{2}{\sqrt3}\times\frac34=\frac{6}{4\sqrt3}=\frac{3}{2\sqrt3}=\frac{\sqrt3}{2}=\sin60^\circ$230^/1+^230^=2/3/1+1/3=2/3/4/3=2/3×34=6/43=3/23=3/2=60^
Answer: (A) $\sin60^\circ$60^, since this expression is the identity $\dfrac{2\tan\theta}{1+\tan^2\theta}=\sin2\theta$2/1+^2=2 evaluated at $\theta=30^\circ$=30^.

(ii) $\tan45^\circ=1$45^=1, so $\tan^245^\circ=1$^245^=1.
$\frac{1-1}{1+1}=\frac{0}{2}=0$1-1/1+1=0/2=0
Answer: (D) $0$0.

(iii) $\sin2A=2\sin A\cos A$2A=2 A A. Setting $2\sin A\cos A=2\sin A$2 A A=2 A gives $\cos A=1$A=1 (dividing by $2\sin A$2 A, valid when $\sin A\ne0$A≠0), which happens at $A=0^\circ$A=0^. Answer: (A) $0^\circ$0^.

(iv) $\tan^230^\circ=\dfrac13$^230^=13.
$\frac{2\tan30^\circ}{1-\tan^230^\circ}=\frac{2/\sqrt3}{1-1/3}=\frac{2/\sqrt3}{2/3}=\frac{2}{\sqrt3}\times\frac32=\frac{3}{\sqrt3}=\sqrt3=\tan60^\circ$230^/1-^230^=2/3/1-1/3=2/3/2/3=2/3×32=3/3=3=60^
Answer: (C) $\tan60^\circ$60^, matching the identity $\dfrac{2\tan\theta}{1-\tan^2\theta}=\tan2\theta$2/1-^2=2 at $\theta=30^\circ$=30^.

$\tan(A+B)=\sqrt3=\tan60^\circ \implies A+B=60^\circ \quad (1)$(A+B)=3=60^ A+B=60^ (1)

$\tan(A-B)=\dfrac{1}{\sqrt3}=\tan30^\circ \implies A-B=30^\circ \quad (2)$(A-B)=1/3=30^ A-B=30^ (2)

Adding (1) and (2): $2A=90^\circ \implies A=45^\circ$2A=90^ A=45^.

Substituting into (1): $B=60^\circ-45^\circ=15^\circ$B=60^-45^=15^.

$A = 45^\circ$A = 45^, $B = 15^\circ$B = 15^.

(i) False. Take $A=B=30^\circ$A=B=30^: $\sin(A+B)=\sin60^\circ=\dfrac{\sqrt3}{2}\approx0.87$(A+B)=60^=3/20.87, but $\sin A+\sin B=\dfrac12+\dfrac12=1$A+ B=12+12=1. These are unequal, so the statement is false in general.

(ii) True. From the standard-values table, as $\theta$ increases from $0^\circ$0^ to $90^\circ$90^, $\sin\theta$ increases from $0$0 to $1$1.

(iii) False. As $\theta$ increases from $0^\circ$0^ to $90^\circ$90^, $\cos\theta$ decreases from $1$1 to $0$0, not increases.

(iv) False. $\sin\theta=\cos\theta$= only at $\theta=45^\circ$=45^ (where both equal $\frac{1}{\sqrt2}$1/2); e.g. at $\theta=30^\circ$=30^, $\sin30^\circ=\frac12\ne\frac{\sqrt3}{2}=\cos30^\circ$30^=12≠3/2=30^.

(v) True. $\cot A=\dfrac{\cos A}{\sin A}$A= A/ A, and $\sin0^\circ=0$0^=0, so $\cot0^\circ$0^ involves division by zero and is therefore not defined.

From $1+\cot^2A=\csc^2A$1+^2A=^2A: $\csc A=\sqrt{1+\cot^2A}$A=√1+^2A, so
$\sin A=\frac{1}{\csc A}=\frac{1}{\sqrt{1+\cot^2A}}$A=1/ A=1/√1+^2A

Since $\tan A=\dfrac{1}{\cot A}$A=1/ A:
$\tan A=\frac{1}{\cot A}$A=1/ A

From $\sec^2A=1+\tan^2A=1+\dfrac{1}{\cot^2A}=\dfrac{\cot^2A+1}{\cot^2A}$^2A=1+^2A=1+1/^2A=^2A+1/^2A:
$\sec A=\frac{\sqrt{1+\cot^2A}}{\cot A}$A=√1+^2A/ A

$\cos A=\frac{1}{\sec A}$A=1/ A

From $\sin^2A=1-\cos^2A=1-\dfrac{1}{\sec^2A}=\dfrac{\sec^2A-1}{\sec^2A}$^2A=1-^2A=1-1/^2A=^2A-1/^2A:
$\sin A=\frac{\sqrt{\sec^2A-1}}{\sec A}$A=√^2A-1/ A

$\tan A=\frac{\sin A}{\cos A}=\frac{\sqrt{\sec^2A-1}}{\sec A}\times\sec A=\sqrt{\sec^2A-1}$A= A/ A=√^2A-1/ A× A=√^2A-1

$\cot A=\frac{1}{\tan A}=\frac{1}{\sqrt{\sec^2A-1}}$A=1/ A=1/√^2A-1

$\csc A=\frac{1}{\sin A}=\frac{\sec A}{\sqrt{\sec^2A-1}}$A=1/ A= A/√^2A-1

(i) $9\sec^2A-9\tan^2A=9(\sec^2A-\tan^2A)=9(1)=9$9^2A-9^2A=9(^2A-^2A)=9(1)=9, using $\sec^2A-\tan^2A=1$^2A-^2A=1. Answer: (B) $9$9.

(ii) Write $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$=/, $\sec\theta=\dfrac{1}{\cos\theta}$=1/, $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$=/, $\csc\theta=\dfrac{1}{\sin\theta}$=1/:
$1+\tan\theta+\sec\theta=\frac{\cos\theta+\sin\theta+1}{\cos\theta},\qquad 1+\cot\theta-\csc\theta=\frac{\sin\theta+\cos\theta-1}{\sin\theta}$1++=++1/, 1+-=+-1/
Multiplying: $\frac{(\sin\theta+\cos\theta+1)(\sin\theta+\cos\theta-1)}{\sin\theta\cos\theta}=\frac{(\sin\theta+\cos\theta)^2-1}{\sin\theta\cos\theta}=\frac{1+2\sin\theta\cos\theta-1}{\sin\theta\cos\theta}=\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}=2$(++1)(+-1)/=(+)^2-1/=1+2-1/=2/=2
Answer: (C) $2$2.

(iii) $(\sec A+\tan A)(1-\sin A)=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)=\frac{(1+\sin A)(1-\sin A)}{\cos A}=\frac{1-\sin^2A}{\cos A}=\frac{\cos^2A}{\cos A}=\cos A$( A+ A)(1- A)=(1/ A+ A/ A)(1- A)=(1+ A)(1- A)/ A=1-^2A/ A=^2A/ A= A
Answer: (D) $\cos A$A.

(iv) $\frac{1+\tan^2A}{1+\cot^2A}=\frac{\sec^2A}{\csc^2A}=\frac{1/\cos^2A}{1/\sin^2A}=\frac{\sin^2A}{\cos^2A}=\tan^2A$1+^2A/1+^2A=^2A/^2A=1/^2A/1/^2A=^2A/^2A=^2A
Answer: (D) $\tan^2A$^2A.