Introduction to Trigonometry — Important Questions

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

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

Answer: (a) $1$

$\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{2}+\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{3}{4}+\dfrac{1}{4}=1$. (This is $\sin(60^\circ+30^\circ)=\sin90^\circ=1$.)

Answer: (c) $1$

$\tan30^\circ=\dfrac{1}{\sqrt3}$ and $\cot60^\circ=\dfrac{1}{\sqrt3}$, so the ratio is $\dfrac{1/\sqrt3}{1/\sqrt3}=1$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Expand each square:

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

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

Add them:

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

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

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

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

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

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

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

Adding the two results:

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

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

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

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