# Chapter-5: Trigonometric Ratios

Q1:

If $$\sin \theta =\frac{\sqrt{3}}{2}$$, find the value of all T-ratios of $$\theta$$.

Sol:

Given:

$$\sin \theta =\frac{\sqrt{3}}{2}$$

Let us draw a $$\bigtriangleup ABC$$ in which $$\angle B=90^{\circ}$$ and $$\angle B= 90^{\circ}=\theta$$.

Then, $$\sin \theta =\frac{BC}{AC}=\frac{\sqrt{3}}{2}$$

Let BC =$$\sqrt{3}k$$

And AC= 2k

where k is positive

By pythagoras theorem, we have

$$AC^{2}=AB^{2}+BC^{2}$$ $$\Rightarrow AB^{2}=AC^{2}-BC^{2}$$ $$\Rightarrow AB^{2}=(2k)^{2}-(\sqrt{3}k)^{2}$$ $$\Rightarrow AB^{2}=(4k^{2}-3k^{2})$$ $$\Rightarrow AB= \sqrt{k^{2}}=k$$

Therefore, $$\sin \theta = \frac{BC}{AC} = \frac{\sqrt{3}k}{2k} =\frac{\sqrt{3}}{2}$$

$$\cos \theta =\frac{AB}{AC}=\frac{k}{2k}=\frac{1}{2}$$ $$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\sqrt{3}}{2}\times \frac{2}{1}=\sqrt{3},$$ $$cosec \theta =\frac{1}{\sin \theta }=\frac{2}{\sqrt{3}},$$

$$\sec \theta =\frac{1}{\cos \theta }=\frac{2}{1}= 2$$ and

$$\cot \theta =\frac{1}{\tan \theta }=\frac{1}{\sqrt{3}}$$

Q2:

If $$\cos \theta =\frac{7}{25}$$, find the value of all T-ratios of $$\theta$$.

Sol:

$$\cos \theta =\frac{\sqrt{7}}{25}$$

Let us draw a $$\bigtriangleup ABC$$ in which $$\angle B = 90^{\circ}$$ and $$\angle BAC= \theta$$

Let AB= 7k and AC= 25k,

Where k is positive

By pythagoras theorem, we have

$$AC^{2}=AB^{2}+BC^{2}$$ $$\Rightarrow BC^{2}=AC^{2}-AB^{2}$$ $$\Rightarrow BC^{2}=(25k)^{2}-(7k)^{2}$$ $$=(625k^{2})-(49k^{2})$$

= $$=576k^{2}$$

$$\Rightarrow BC= \sqrt{576k^{2}}=24k$$

Therefore, $$\sin \theta = \frac{BC}{AC} = \frac{24k}{25k} = \frac{24}{25},$$

$$\cos \theta =\frac{7}{5}$$    ( given )

$$\tan \theta =\frac{\sin \theta }{\cos \theta }= \frac{24}{25}\times \frac{25}{7}=\frac{24}{7}$$ $$cosec \theta =\frac{1}{\sin \theta }=\frac{25}{24}$$

$$\sec \theta =\frac{1}{\cos \theta }=\frac{25}{7}$$ and

$$\cot \theta =\frac{1}{\tan \theta }=\frac{7}{24}$$

Q3:

If $$\tan \theta =\frac{15}{8}$$, find the value of all T-ratios of $$\theta$$.

Sol:

Given:

$$\tan \theta =\frac{BC}{AB} = \frac{15}{8}$$

Let AB=15k and AC=8k,

Where k is positive

Let us draw a $$\bigtriangleup ABC$$ in which $$\angle B=90^{\circ}$$ and $$\angle BAC=90^{\circ} \; and \; \angle BAC=\theta$$

By pythagoras theorem, we have

$$AC^{2}=AB^{2}+BC^{2}$$ $$=(8k)^{2}+(15)k^{2}$$ $$=(64k^{2})+(225k^{2})$$

= $$=289k^{2}$$

AC=17k

Therefore, $$\sin \theta= \frac{BC}{AC} = \frac{15k} {17k} = \frac{15} {17}$$

$$\cos \theta= \frac{AB} {AC} = \frac{8k} {17k} = \frac{8} {17}$$

$$\tan \theta= \frac{15} {8}$$   (given)

$$cosec \theta = \frac{1}{\sin \theta } = \frac{17}{15}$$ $$\sec \theta= \frac{1}{\cos \theta } = \frac{17}{8}$$

and $$\cot \theta= \frac{1}{\tan \theta } = \frac{8}{15}$$

Q4:

If $$\cot \theta = 2$$, find the value of all T-ratios of $$\theta$$.

Sol:

Given:

$$\cot \theta =\frac{AB}{BC}=\frac{2}{1}$$

Let AB = 2k and AC = 1k,

Where k is positive

Let us draw a $$\bigtriangleup ABC$$ in which $$\angle B=90^{\circ}$$ and $$\angle BAC = \theta$$

By pythagoras theorem, we have

$$AC^{2} = AB^{2}+BC^{2}$$ $$AC^{2} = (2k)^{2}+(k)^{2}$$ $$AC^{2} = 4k^{2}+k^{2}=5k^{2}$$

Therefore, $$AC = \sqrt{5k^{2}} = \sqrt{5}k$$

$$\sin \theta = \frac{BC}{AC} = \frac{k}{\sqrt{5}k} = \frac{1}{\sqrt{5}}$$ $$\cos \theta = \frac{AB}{AC} = \frac{2k}{\sqrt{5}k} = \frac{2} {\sqrt{5}}$$ $$\tan \theta = \frac{1}{\cot \theta } = \frac{1}{2}$$ $$\cot \theta = 2$$ $$cosec \theta = \frac{1}{\sin \theta } = \sqrt{5}$$ $$sec \theta = \frac{1}{\cos \theta } = \frac{\sqrt{5}} {2}$$

Q5:

If $$cosec \theta = \sqrt{10}$$, find the value of all T-ratios of $$\theta$$.

Sol:

Given:

$$cosec \theta =\frac{AC}{BC}=\frac{\sqrt{10}}{1}$$

Let $$AB = \sqrt{10}k$$ and AC = 1k,

Where k is positive

Let us draw a $$\bigtriangleup ABC$$ in which $$\angle B=90^{\circ}$$ and $$\angle BAC = \theta$$

By pythagoras theorem, we have

$$AC^{2} = AB^{2}+BC^{2}$$ $$AB^{2} = AC^{2} – BC^{2}$$ $$AB^{2} = (\sqrt{10}k)^{2} – (k)^{2}$$ $$AB^{2} = 10k^{2} – k^{2} = 9k^{2}$$

Therefore, $$AB = \sqrt{9k^{2}} = 3k$$

$$\sin \theta = \frac{BC}{AC} = \frac{1}{\sqrt{10}}$$ $$\cos \theta = \frac{AB}{AC} = \frac{3k}{\sqrt{10}k} = \frac{3} {\sqrt{10}}$$

$$cosec \theta = \sqrt{10}$$         (given)

$$sec \theta = \frac{1}{\cos \theta } = \frac{\sqrt{10}} {3}$$ $$\tan \theta = \frac{\sin \theta }{\cos \theta }= \left ( \frac{1}{\sqrt{10}} \times \frac{\sqrt{10}}{3} \right ) = \frac{1}{3}$$ $$\cot \theta = \frac{1 }{\tan \theta }= 3$$

Q6:

If $$\sin \theta = \frac{a^{2} – b^{2}}{a^{2} + b^{2}}$$, find the values of all T-ratio of $$\theta$$.

Sol:

Given:

We have,

$$\sin \theta = \frac{a^{2} – b^{2}}{a^{2} + b^{2}}$$

As,

$$\cos ^{2}\theta =1-\sin ^{2}\theta$$ $$= 1- \left ( \frac{a^{2}-b^{2}}{a^{2}+b^{2}} \right )^{2}$$ $$= \frac {(a^{2}+b^{2})^{2} – (a^{2} – b^{2})^{2}}{{(a^{2}+b^{2}}) ^{2}}$$ $$= \frac {[(a^{2}+b^{2}) – (a^{2} – b^{2})][(a^{2}+b^{2}) + (a^{2} – b^{2})]}{{(a^{2}+b^{2}}) ^{2}}$$ $$= \frac {[a^{2}+b^{2} – a^{2} + b^{2}][a^{2}+b^{2} + a^{2} – b^{2}]}{{(a^{2}+b^{2}}) ^{2}}$$ $$= \frac {[2b^{2}][2a^{2}]}{{(a^{2}+b^{2}}) ^{2}}$$ $$\cos ^{2} \theta = \frac {4a^{2}b^{2}}{{(a^{2}+b^{2}}) ^{2}}$$ $$\cos \theta = \sqrt{\frac {4a^{2}b^{2}}{{(a^{2}+b^{2}}) ^{2}}}$$ $$\cos \theta = \frac {2ab}{{(a^{2}+b^{2}}) }$$

Also,

$$\tan \theta = \frac {\sin \theta}{\cos \theta }$$ $$\tan \theta = \frac {\frac{a^{2}-b^{2}}{a^{2}+b^{2}}}{\frac{2ab}{a^{2}+b^{2}}}$$ $$= \frac{a^{2} – b^{2}}{2ab}$$ $$cosec\; \theta = \frac{1}{\sin \theta}$$ $$= \frac{1}{\frac{a^{2} – b^{2}}{a^{2} + b^{2}}}$$ $$= \frac{a^{2} + b^{2}}{a^{2} – b^{2}}$$

Also,

$$\sec \theta = \frac{1}{\cos \theta}$$ $$= \frac{1}{\frac {2ab}{{(a^{2}+b^{2}}) }}$$ $$= \frac{(a^{2}+b^{2})}{2ab}$$

And

$$\cot \theta =\frac{1}{\tan \theta }$$ $$= \frac{1}{\frac{a^{2}-b^{2}}{2ab}}$$ $$= \frac{2ab}{a^{2}-b^{2}}$$

Q7:

If $$15\cot \theta = 8$$, find the value of $$\sin \theta \; and \; \sec \theta$$.

Sol:

Given:

$$\cot \theta =\frac{AB}{BC}=\frac{15}{8}$$

Let AB = 15k and BC = 8k,

Where k is positive

Let us draw a $$\bigtriangleup ABC$$ in which $$\angle B=90^{\circ}$$ and $$\angle BAC = \theta$$

By pythagoras theorem, we have

$$AC^{2} = AB^{2} + BC^{2}$$ $$AC^{2} = (15k)^{2} + (8k)^{2}$$ $$AC^{2} = (225 k)^{2} + 64k^{2}$$ $$AC^{2} = 289k^{2} = 17k^{2}$$

Therefore, $$AC = \sqrt{289k^{2}} = 17k$$

$$\sin \theta = \frac{BC}{AC} = \frac{8k}{17k} \frac{8} {17}$$ $$\cos \theta = \frac{AB}{AC} = \frac{15k}{17k} = \frac{15} {17}$$

Therefore, $$\frac{(2 + 2sin\theta)(1- \sin \theta)}{(1+ \cos \theta) (2-2 \cos \theta)} = \frac{(2+2\times \frac{8}{17}) (1-\frac{8}{17})} {(1+\frac{15}{17}) (2-2\times \frac{15} {17})}$$

$$\frac{\frac{50}{17}\times \frac{9}{17}}{\frac{32}{17}\times \frac{4}{17}}=\frac{50\times 9}{32\times 4}=\frac{225}{64}$$

Q8:

If sin A = $$\frac{9}{41}$$, find the values of cos A and tan A.

Sol:

We have sin A = $$\frac{9}{41}$$

As,

$$cos^{2}A=1-sin^{2}A$$ $$=1-\left ( \frac{9}{41} \right )^{2}$$ $$=1-\frac{81}{1681}$$ $$=\frac{1681-81}{1681}$$ $$\Rightarrow cos^{2}A=\frac{1600}{1681}$$ $$\Rightarrow cosA=\sqrt{\frac{1600}{1681}}$$ $$\Rightarrow cosA=\frac{40}{41}$$

Also.

$$tanA=\frac{sinA}{cosA}$$

= $$\frac{\left ( \frac{9}{41} \right )}{\left ( \frac{40}{41} \right )}$$

$$=\frac{9}{40}$$

Q9:

If $$\cos \theta = 0.6$$, show that $$(5\sin \theta -3\tan \theta) = 0$$.

Sol:

Given:

$$\cos \theta = 0.6 = \frac{6} {10} = \frac{3} {5}$$

Let us draw a $$\bigtriangleup ABC$$ in which $$\angle B =90^{\circ} \; and \; \angle BCA = \theta$$

Then, $$\cos \theta = \frac{AB}{AC}= \frac{3}{5}$$

Let AB = 3k

and AC = 5k

Where k is positive.

By pythagoras theorem, we have

$$AC^{2} = AB^{2} + BC^{2}$$ $$BC^{2}= AC^{2} – AB^{2}$$ $$=\left [ (5k)^{2} – (3k)^{2} \right ] = 16k^{2}$$ $$BC^{2} = 16k^{2}$$ $$BC= 4k$$ $$\sin \theta= \frac{AB}{AC} = \frac{4k} {5k} = \frac{4} {5}$$ $$\cos \theta =\frac{3} {5}$$ $$\tan \theta = \frac{\sin \theta }{\cos \theta }= \frac{4}{5}\times \frac{5}{3}= \frac{4}{3}$$ $$\Rightarrow (5\sin \theta -3\tan \theta) = 5\times \frac{4}{5} – 3\times \frac{4}{3} = 0$$

Hence, $$(5\sin \theta -3\tan \theta) = 0$$

Q10:

If $$cosec \theta = 2$$, show that $$\cot \theta +\frac{\sin \theta }{1+\cos \theta } = 2$$

Sol:

Given:

$$cosec \theta =\frac{AC}{BC}=\frac{2}{1}$$

Let $$AC = 2k$$ and BC = 1k,

Where k is positive

Let us draw a $$\bigtriangleup ABC$$ in which $$\angle B=90^{\circ}$$ and $$\angle BAC = \theta$$

By pythagoras theorem, we have

$$AC^{2} = AB^{2}+BC^{2}$$ $$AB^{2} = AC^{2} – BC^{2}$$ $$AB^{2} = (2k)^{2} – (k)^{2}$$ $$AB^{2} = 4k^{2} – k^{2} = 3k^{2}$$

Therefore, $$AB = \sqrt{3k^{2}} = \sqrt{3} k$$

$$\sin \theta = \frac{BC}{AC} = \frac{1}{2}$$ $$\cos \theta = \frac{AB}{AC} = \frac{ \sqrt{3}k} {2k} = \frac{\sqrt{3}} {2}$$ $$\cot \theta = \frac{\sin \theta } {\cos \theta } = \frac{\sqrt{3}} {2} \times \frac{2}{1} = \sqrt{3}$$ $$\Rightarrow \left [ \cot \theta +\frac{\sin \theta }{1+\cos \theta } \right ]= \left [ \sqrt{3} + \frac{\frac{1}{2}}{1+\frac{3}{2}} \right ]$$ $$= \left [ \sqrt{3}+\frac{1}{2+\sqrt{3}} \right ]= \left [ \frac{2\sqrt{3}+3+1}{2+\sqrt{3}} \right ]$$ $$= \left [ \frac{2\sqrt{3}+4}{2+\sqrt{3}} \right ] = 2\left [ \frac{\sqrt{3}+2}{2+\sqrt{3}} \right ] = 2$$

Hence, $$\cot \theta +\frac{\sin \theta }{1+\cos \theta } =2$$

Q11:

If $$\tan \theta = \frac{1}{\sqrt{7}}$$, $$\frac{(cosec^{2}\theta – \sec ^{2} \theta )}{(cosec^{2}\theta + \sec ^{2} \theta )} = \frac{3}{4}$$ .

Sol:

Given:

$$\tan \theta =\frac{BC}{AB}= \frac{1} {\sqrt{7}}$$

Let BC = 1k and AB = $$\sqrt{7}k$$,

Where k is positive

Let us draw a $$\bigtriangleup ABC$$ in which $$\angle B=90^{\circ}\; and \; \angle BAC=\theta$$

By pythagoras theorem, we have

$$AC^{2} = AB^{2} + BC^{2}$$ $$= (\sqrt{7}k)^{2}+(k)^{2}$$ $$= (7k^{2})+(k^{2})$$ $$= 8k^{2}$$

AC = $$2\sqrt{2}k$$

Therefore, $$cosec \theta= \frac{AC}{BC} = \frac{2\sqrt{2}k} {1k} = 2\sqrt{2}$$

$$\sec \theta= \frac{AC} {AB} = \frac{2\sqrt{2}k}{\sqrt{7}k} = \frac{2\sqrt{2}}{\sqrt{7}}$$ $$\Rightarrow \frac{(cosec^{2}\theta – \sec ^{2} \theta )}{(cosec^{2}\theta + \sec ^{2} \theta )} = \begin{bmatrix} \frac{(2\sqrt{2})^{2}-(\frac{2\sqrt{2}}{\sqrt{7}})^{2}}{(2\sqrt{2})^{2} + (\frac{2\sqrt{2}}{\sqrt{7}})^{2}} \end{bmatrix}$$

= $$\left ( \frac{8 – \frac{8}{7}}{8 + \frac{8}{7}} \right )=\frac{\left ( \frac{48}{7} \right )}{\left ( \frac{64}{7} \right )}=\frac{48}{64} = \frac{3}{4}$$

$$\Rightarrow \frac{(cosec^{2}\theta – \sec ^{2} \theta )}{(cosec^{2}\theta + \sec ^{2} \theta )} = \frac{3}{4}$$

Q12:

If $$tan\Theta =\frac{20}{21}$$, show that $$\frac{\left ( 1-sin\Theta +cos\Theta \right )}{\left ( 1+sin\Theta +cos\Theta \right )}=\frac{3}{7}$$

Sol:

Given:

$$\tan \theta = \frac{20}{21}= \frac{20k}{21k}$$

Let us draw a $$\bigtriangleup ABC$$ in which $$\angle B = 90^{\circ} \; and \; \angle A = \theta$$

By pythagoras theorem, we have

$$AC^{2} = AB^{2} + BC^{2} = (21k)^{2} + (20k)^{2}$$ $$= 441k^{2} + 400^{2}$$ $$= 841k^{2}$$

Therefore, $$AC = 29k$$

$$\sin \theta = \frac{BC}{AC} = \frac{20k}{29k} = \frac{20}{29}$$ $$\cos \theta = \frac{AB}{AC} = \frac{21k}{29k} = \frac{21}{29}$$

L.H.S. = $$= \frac{1- \sin \theta +\cos \theta }{1+ \sin \theta +\cos \theta } = \frac{1 – \frac{20}{29} + \frac{21}{29}}{1 + \frac{20}{29} + \frac{21}{29}} = \frac{\frac{29-20+21}{29}}{\frac{29+21+21}{29}}$$

= $$\frac{20}{70} = \frac{3}{7}$$ = R.H.S.

Q13:

$$sec\Theta =\frac{5}{4},\:show\:that\:\frac{\left ( sin\Theta -2cos\Theta \right )}{tan\Theta -cot\Theta }=\frac{12}{7}$$

Sol:

We have,

$$sec\Theta =\frac{5}{4}$$ $$\Rightarrow \frac{1}{cos\Theta }=\frac{5}{4}$$ $$\Rightarrow cos\Theta =\frac{4}{5}$$

Also,

$$sin^{2}\Theta =1-cos^{2}\Theta$$ $$=1-\left ( \frac{4}{5} \right )^{2}$$ $$=1-\frac{16}{25}$$ $$=\frac{9}{25}$$ $$\Rightarrow sin\Theta =\frac{3}{5}$$

Now,

$$L.H.S=\frac{\left ( sin\Theta -2cos\Theta \right )}{\left ( tan\Theta -cot\Theta \right ) }$$ $$=\frac{\left ( sin\Theta -2cos\Theta \right )}{\left ( \frac{sin\Theta }{cos\Theta }-\frac{cos\Theta }{sin\Theta } \right )}$$ $$=\frac{\left ( sin\Theta -2cos\Theta \right )}{\left ( \frac{sin^{2}\Theta -cos^{2}\Theta }{sin\Theta cos\Theta } \right )}$$ $$=\frac{sin\Theta cos\Theta \left ( sin\Theta -2cos\Theta \right )}{sin^{2}\Theta -cos^{2}\Theta }$$ $$=\frac{\frac{3}{5}\times \frac{4}{5}\left ( \frac{3}{5}-2\times \frac{4}{5} \right )}{\left ( \frac{3}{5} \right )^{2}-\left ( \frac{4}{5} \right )^{2}}$$ $$=\frac{\frac{12}{25}\left ( \frac{3}{5}-\frac{8}{5} \right )}{\left ( \frac{9}{25}-\frac{16}{25} \right )}$$ $$=\frac{\frac{12}{5}\times \left ( \frac{-5}{5} \right )}{\left ( \frac{-7}{25} \right )}$$ $$=\frac{12}{7}$$

R.H.S

Q14:

If $$cot\Theta =\frac{3}{4},\:show\:that\:\sqrt{\frac{sec\Theta -cosec\Theta }{sec\Theta +cosec\Theta }}=\frac{1}{\sqrt{7}}$$

Sol:

L.H.S = $$\sqrt{\frac{sec\Theta -cosec\Theta }{sec\Theta +cosec\Theta }}$$

$$=\sqrt{\frac{\left ( \frac{1}{cos\Theta }-\frac{1}{sin\Theta } \right )}{\left ( \frac{1}{cos\Theta }+\frac{1}{sin\Theta } \right )}}$$ $$=\sqrt{\frac{\left ( \frac{sin\Theta -cos\Theta }{sin\Theta cos\Theta } \right )}{\left ( \frac{sin\Theta +cos\Theta }{sin\Theta cos\Theta } \right )}}$$ $$=\sqrt{\frac{\left ( \frac{sin\Theta -cos\Theta }{sin\Theta } \right )}{\left ( \frac{sin\Theta +cos\Theta }{sin\Theta } \right )}}$$ $$=\sqrt{\frac{\left ( \frac{sin\Theta }{sin\Theta }-\frac{cos\Theta }{sin\Theta } \right )}{\left ( \frac{sin\Theta }{sin\Theta }+\frac{cos\Theta }{sin\Theta } \right )}}$$ $$=\sqrt{\frac{1-cot\Theta }{1+cot\Theta }}$$ $$=\sqrt{\frac{\left ( 1-\frac{3}{4} \right )}{\left ( 1+\frac{3}{4} \right )}}$$ $$=\sqrt{\frac{\left ( \frac{1}{4} \right )}{\left ( \frac{7}{4} \right )}}$$ $$=\sqrt{\frac{1}{7}}$$ $$=\frac{1}{\sqrt{7}}$$

= R.H.S

Q15:

If $$sin\Theta =\frac{3}{4},\:show\:that\:\sqrt{\frac{cosec^{2}\Theta -cot^{2}\Theta }{sec^{2}\Theta -1}}=\frac{\sqrt{7}}{3}$$

Sol:

L.H.S = $$\sqrt{\frac{cosec^{2}\Theta -cot^{2}\Theta }{sec^{2}\Theta -1}}$$

$$=\sqrt{\frac{1}{tan^{2}\Theta }}$$ $$=\sqrt{cot^{2}\Theta }$$ $$=cot\Theta$$ $$=\sqrt{cosec^{2}\Theta -1 }$$ $$=\sqrt{\left ( \frac{1}{sin\Theta }^{2}-1 \right )}$$ $$=\sqrt{\left ( \frac{1}{\left ( \frac{3}{4} \right )} \right )^{2}-1}$$ $$=\sqrt{\left ( \frac{4}{3} \right )^{2}-1}$$ $$=\sqrt{\frac{16}{9}-1}$$ $$=\sqrt{\frac{16-9}{9}}$$ $$=\sqrt{\frac{7}{9}}$$ $$=\frac{\sqrt{7}}{3}$$

= R.H.S