RS Aggarwal Class 10 Solutions Trigonometric Ratios

RS Aggarwal Class 10 Solutions Chapter 5

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 \(\angle B =90^{\circ} \; and \; \angle BCA = \theta\)

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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


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