Introduction to Trigonometry — CBSE Class 10 Maths Important Questions

Answer: (a) $\dfrac{4}{5}$4/5

Since $\theta$θ is acute, $\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}$cos θ=√(1-sin²θ)=1-9/25=16/25=4/5.

Answer: (a) $1$1

$\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{2}+\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{3}{4}+\dfrac{1}{4}=1$√3/2× √3/2+1/2× 1/2=3/4+1/4=1.

Answer: (c) $1$1

$\tan30^\circ=\dfrac{1}{\sqrt3}$tan 30°=1/√3 and $\cot60^\circ=\dfrac{1}{\sqrt3}$cot 60°=1/√3, so the ratio is $\dfrac{1/\sqrt3}{1/\sqrt3}=1$1/√3/1/√3=1.

Answer: (b) $A=45^\circ,\ B=45^\circ$A=45°, B=45°

$\sin(A+B)=1\Rightarrow A+B=90^\circ$sin (A+B)=1⇒ A+B=90° and $\cos(A-B)=1\Rightarrow A-B=0^\circ$cos (A-B)=1⇒ A-B=0°. Solving, $A=B=45^\circ$A=B=45°.

Answer: (a) Both A and R are true and R is the correct explanation of A.

$\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}$sin θ=opposite/hypotenuse, and the opposite side is always shorter than the hypotenuse, so $\sin\theta<1$sin θ<1 for every acute angle. Since $\dfrac{4}{3}>1$4/3>1, $\sin\theta$sin θ can never equal $\dfrac{4}{3}$4/3.

From $1+\cot^2 A=\csc^2 A$1+cot² A=csc² A, $\csc A=\sqrt{1+\cot^2 A}$csc A=√(1+cot² A), so $\sin A=\dfrac{1}{\csc A}=\dfrac{1}{\sqrt{1+\cot^2 A}}$sin A=1/(csc A)=1/(√(1+cot² A)).

Since $\tan A=\dfrac{1}{\cot A}$tan A=1/(cot A), $\sec^2 A=1+\tan^2 A=1+\dfrac{1}{\cot^2 A}=\dfrac{\cot^2 A+1}{\cot^2 A}$sec² A=1+tan² A=1+1/(cot² A)=(cot² A+1)/(cot² A).

So $\sec A=\dfrac{\sqrt{1+\cot^2 A}}{\cot A}$sec A=(√(1+cot² A))/(cot A).

$\sqrt3\,\tan\theta=3\sin\theta\Rightarrow \sqrt3\cdot\dfrac{\sin\theta}{\cos\theta}=3\sin\theta$√3 tan θ=3 sin θ ⇒ √3· (sin θ)/(cos θ)=3 sin θ.

Since $\theta$θ is acute, $\sin\theta\ne0$sin θ ≠0, so $\dfrac{\sqrt3}{\cos\theta}=3\Rightarrow \cos\theta=\dfrac{1}{\sqrt3}$√3/(cos θ)=3⇒ cos θ=1/√3.

Then $\cos^2\theta=\dfrac{1}{3}$cos²θ=1/3 and $\sin^2\theta=1-\dfrac{1}{3}=\dfrac{2}{3}$sin²θ=1-1/3=2/3.

Therefore $\sin^2\theta-\cos^2\theta=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}$sin²θ-cos²θ=2/3-1/3=1/3.

$4\tan\theta=3\Rightarrow \tan\theta=\dfrac{3}{4}$4 tan θ=3⇒ tan θ=3/4. This corresponds to a right triangle with opposite $=3$=3, adjacent $=4$=4, hypotenuse $=\sqrt{3^2+4^2}=5$=√(3²+4²)=5.

So $\sin\theta=\dfrac{3}{5}$sin θ=3/5 and $\cos\theta=\dfrac{4}{5}$cos θ=4/5.

Numerator $=4\cdot\dfrac{3}{5}-\dfrac{4}{5}+1=\dfrac{12}{5}-\dfrac{4}{5}+\dfrac{5}{5}=\dfrac{13}{5}$=4· 3/5-4/5+1=12/5-4/5+5/5=13/5.

Denominator $=4\cdot\dfrac{3}{5}+\dfrac{4}{5}+1=\dfrac{12}{5}+\dfrac{4}{5}+\dfrac{5}{5}=\dfrac{21}{5}$=4· 3/5+4/5+1=12/5+4/5+5/5=21/5.

Therefore the value $=\dfrac{13/5}{21/5}=\dfrac{13}{21}$=13/5/21/5=13/21.

Taking the LCM $(1+\sin A)\cos A$(1+sin A)cos A:

$\text{LHS}=\dfrac{\cos^2 A+(1+\sin A)^2}{(1+\sin A)\cos A}$LHS=(cos² A+(1+sin A)²)/((1+sin A)cos A).

Expand the numerator: $\cos^2 A+1+2\sin A+\sin^2 A=(\sin^2 A+\cos^2 A)+1+2\sin A=1+1+2\sin A=2(1+\sin A)$cos² A+1+2 sin A+sin² A=(sin² A+cos² A)+1+2 sin A=1+1+2 sin A=2(1+sin A).

So $\text{LHS}=\dfrac{2(1+\sin A)}{(1+\sin A)\cos A}=\dfrac{2}{\cos A}=2\sec A=\text{RHS}$LHS=(2(1+sin A))/((1+sin A)cos A)=2/(cos A)=2 sec A=RHS. Hence proved.

Write everything in terms of $\tan\theta$tan θ. Since $\cot\theta=\dfrac{1}{\tan\theta}$cot θ=1/(tan θ):

$\dfrac{\tan\theta}{1-\cot\theta}=\dfrac{\tan\theta}{1-\frac{1}{\tan\theta}}=\dfrac{\tan^2\theta}{\tan\theta-1}$(tan θ)/(1-cot θ)=tan θ1-1/(tan θ)=(tan²θ)/(tan θ-1).

$\dfrac{\cot\theta}{1-\tan\theta}=\dfrac{1/\tan\theta}{1-\tan\theta}=\dfrac{1}{\tan\theta(1-\tan\theta)}=\dfrac{-1}{\tan\theta(\tan\theta-1)}$(cot θ)/(1-tan θ)=(1/tan θ)/(1-tan θ)=1/(tan θ(1-tan θ))=-1/(tan θ(tan θ-1)).

Adding: $\dfrac{\tan^2\theta}{\tan\theta-1}-\dfrac{1}{\tan\theta(\tan\theta-1)}=\dfrac{\tan^3\theta-1}{\tan\theta(\tan\theta-1)}$(tan²θ)/(tan θ-1)-1/(tan θ(tan θ-1))=(tan³θ-1)/(tan θ(tan θ-1)).

Using $\tan^3\theta-1=(\tan\theta-1)(\tan^2\theta+\tan\theta+1)$tan³θ-1=(tan θ-1)(tan²θ+tan θ+1):

$=\dfrac{\tan^2\theta+\tan\theta+1}{\tan\theta}=\tan\theta+1+\cot\theta=1+(\tan\theta+\cot\theta)$=(tan²θ+tan θ+1)/(tan θ)=tan θ+1+cot θ=1+(tan θ+cot θ).

Now $\tan\theta+\cot\theta=\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\dfrac{1}{\sin\theta\cos\theta}=\sec\theta\,\csc\theta$tan θ+cot θ=(sin²θ+cos²θ)/(sin θ cos θ)=1/(sin θ cos θ)=sec θ csc θ.

Hence LHS $=1+\sec\theta\,\csc\theta=$=1+sec θ csc θ= RHS. Proved.

Since $\sec^2\theta-\tan^2\theta=1$sec²θ-tan²θ=1, we have $(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1$(sec θ+tan θ)(sec θ-tan θ)=1, so $\sec\theta-\tan\theta=\dfrac{1}{2}$sec θ-tan θ=1/2.

Adding the two equations: $2\sec\theta=2+\dfrac{1}{2}=\dfrac{5}{2}$2 sec θ=2+1/2=5/2, so $\sec\theta=\dfrac{5}{4}$sec θ=5/4.

Subtracting: $2\tan\theta=2-\dfrac{1}{2}=\dfrac{3}{2}$2 tan θ=2-1/2=3/2, so $\tan\theta=\dfrac{3}{4}$tan θ=3/4.

$\sin\theta=\dfrac{\tan\theta}{\sec\theta}=\dfrac{3/4}{5/4}=\dfrac{3}{5}$sin θ=(tan θ)/(sec θ)=3/4/5/4=3/5.

(Check: $1+\left(\dfrac{3}{4}\right)^2=\dfrac{25}{16}=\left(\dfrac{5}{4}\right)^2$1+(3/4)²=25/16=(5/4)².)

Expand each square:

$(\sin A+\csc A)^2=\sin^2 A+2\sin A\csc A+\csc^2 A=\sin^2 A+2+\csc^2 A$(sin A+csc A)²=sin² A+2 sin A csc A+csc² A=sin² A+2+csc² A (since $\sin A\csc A=1$sin A csc A=1).

$(\cos A+\sec A)^2=\cos^2 A+2\cos A\sec A+\sec^2 A=\cos^2 A+2+\sec^2 A$(cos A+sec A)²=cos² A+2 cos A sec A+sec² A=cos² A+2+sec² A (since $\cos A\sec A=1$cos A sec A=1).

Add them:

$\text{LHS}=(\sin^2 A+\cos^2 A)+4+(\csc^2 A+\sec^2 A)$LHS=(sin² A+cos² A)+4+(csc² A+sec² A).

Use $\sin^2 A+\cos^2 A=1$sin² A+cos² A=1, $\csc^2 A=1+\cot^2 A$csc² A=1+cot² A and $\sec^2 A=1+\tan^2 A$sec² A=1+tan² A:

$=1+4+(1+\cot^2 A)+(1+\tan^2 A)=7+\tan^2 A+\cot^2 A=\text{RHS}$=1+4+(1+cot² A)+(1+tan² A)=7+tan² A+cot² A=RHS. Hence proved.

Rationalise the first radical by multiplying numerator and denominator inside by $(1+\sin A)$(1+sin A):

$\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sqrt{\dfrac{(1+\sin A)^2}{(1-\sin A)(1+\sin A)}}=\sqrt{\dfrac{(1+\sin A)^2}{1-\sin^2 A}}=\sqrt{\dfrac{(1+\sin A)^2}{\cos^2 A}}=\dfrac{1+\sin A}{\cos A}$(1+sin A)/(1-sin A)=((1+sin A)²)/(1-sin A)(1+sin A)=((1+sin A)²)/(1-sin² A)=((1+sin A)²)/(cos² A)=(1+sin A)/(cos A),

since $A$A is acute so $\cos A>0$cos A>0 and $1+\sin A>0$1+sin A>0.

Similarly $\sqrt{\dfrac{1-\sin A}{1+\sin A}}=\dfrac{1-\sin A}{\cos A}$(1-sin A)/(1+sin A)=(1-sin A)/(cos A).

Adding the two results:

$\text{LHS}=\dfrac{1+\sin A}{\cos A}+\dfrac{1-\sin A}{\cos A}=\dfrac{(1+\sin A)+(1-\sin A)}{\cos A}=\dfrac{2}{\cos A}=2\sec A=\text{RHS}$LHS=(1+sin A)/(cos A)+(1-sin A)/(cos A)=(1+sin A)+(1-sin A)/(cos A)=2/(cos A)=2 sec A=RHS. Hence proved.

(i) By Pythagoras, $PR=\sqrt{PQ^2+QR^2}=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$PR=√(PQ²+QR²)=√(5²+12²)=√(25+144)=√169=13 units.

(ii) With respect to $\angle P$∠ P, the opposite side is $QR=12$QR=12 and the adjacent side is $PQ=5$PQ=5. So $\sin P=\dfrac{QR}{PR}=\dfrac{12}{13}$sin P=QR/PR=12/13 and $\cos P=\dfrac{PQ}{PR}=\dfrac{5}{13}$cos P=PQ/PR=5/13.

(iii) $\tan P=\dfrac{QR}{PQ}=\dfrac{12}{5}$tan P=QR/PQ=12/5, so $1+\tan^2 P=1+\dfrac{144}{25}=\dfrac{169}{25}$1+tan² P=1+144/25=169/25. Also $\sec P=\dfrac{PR}{PQ}=\dfrac{13}{5}$sec P=PR/PQ=13/5, so $\sec^2 P=\dfrac{169}{25}$sec² P=169/25. Since both equal $\dfrac{169}{25}$169/25, the identity $1+\tan^2 P=\sec^2 P$1+tan² P=sec² P is verified.