Introduction to Trigonometry — CBSE Class 10 Maths Important Questions

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

Since $\theta$ is acute, $\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}$=√1-^2=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. (This is $\sin(60^\circ+30^\circ)=\sin90^\circ=1$(60^+30^)=90^=1.)

Answer: (c) $1$1

$\tan30^\circ=\dfrac{1}{\sqrt3}$30^=1/3 and $\cot60^\circ=\dfrac{1}{\sqrt3}$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^

$\sin(A+B)=1\Rightarrow A+B=90^\circ$(A+B)=1 A+B=90^ and $\cos(A-B)=1\Rightarrow A-B=0^\circ$(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.

$\sin^2 30^\circ+\cos^2 30^\circ=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt3}{2}\right)^2=\dfrac{1}{4}+\dfrac{3}{4}=1$^2 30^+^2 30^=(1/2)^2+(3/2)^2=1/4+3/4=1, which is exactly the identity in R evaluated at $\theta=30^\circ$=30^.

By Pythagoras, $AC=\sqrt{AB^2+BC^2}=\sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=25$AC=√AB^2+BC^2=√24^2+7^2=√576+49=√625=25 cm.

$\sin A=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{BC}{AC}=\dfrac{7}{25}$A=opposite/hypotenuse=BC/AC=7/25 and $\cos A=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{AB}{AC}=\dfrac{24}{25}$A=adjacent/hypotenuse=AB/AC=24/25.

$\sqrt3\,\tan\theta=3\sin\theta\Rightarrow \sqrt3\cdot\dfrac{\sin\theta}{\cos\theta}=3\sin\theta$3\,=3 3·/=3.

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

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

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

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

So $\sin\theta=\dfrac{3}{5}$=3/5 and $\cos\theta=\dfrac{4}{5}$=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+ A) A:

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

Expand the numerator: $\cos^2 A+1+2\sin A+\sin^2 A=(\sin^2 A+\cos^2 A)+1+2\sin A=1+1+2\sin A=2(1+\sin A)$^2 A+1+2 A+^2 A=(^2 A+^2 A)+1+2 A=1+1+2 A=2(1+ 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+ A)/(1+ A) A=2/ A=2 A=RHS. Hence proved.

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

$\dfrac{\tan\theta}{1-\cot\theta}=\dfrac{\tan\theta}{1-\frac{1}{\tan\theta}}=\dfrac{\tan^2\theta}{\tan\theta-1}$/1-=1-1/=^2/-1.

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

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

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

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

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

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

Expand each square:

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

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

Add them:

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

Use $\sin^2 A+\cos^2 A=1$^2 A+^2 A=1, $\csc^2 A=1+\cot^2 A$^2 A=1+^2 A and $\sec^2 A=1+\tan^2 A$^2 A=1+^2 A:

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

Rationalise the first radical by multiplying numerator and denominator inside by $(1+\sin A)$(1+ 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+ A/1- A=(1+ A)^2/(1- A)(1+ A)=(1+ A)^2/1-^2 A=(1+ A)^2/^2 A=1+ A/ A,

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

Similarly $\sqrt{\dfrac{1-\sin A}{1+\sin A}}=\dfrac{1-\sin A}{\cos A}$1- A/1+ A=1- A/ 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+ A/ A+1- A/ A=(1+ A)+(1- A)/ A=2/ A=2 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=√PQ^2+QR^2=√5^2+12^2=√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 $\sin P=\dfrac{QR}{PR}=\dfrac{12}{13}$P=QR/PR=12/13 and $\cos P=\dfrac{PQ}{PR}=\dfrac{5}{13}$P=PQ/PR=5/13.

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