Well-structured S Chand Class 11 Maths Solutions Chapter 4 Trigonometrical Functions Ex 4(c) facilitate a deeper understanding of mathematical principles.
S Chand Class 11 ICSE Maths Solutions Chapter 4 Trigonometrical Functions Ex 4(c)
Question 1.
If cot θ = \(\frac { 4 }{ 3 }\), find the values of other t-ratios of θ.
Solution:
Question 2.
If cos A = 0.6, find the values of 5 sin A – 3 tan A.
Solution:
Given cos A = 0.6 = \(\frac { 6 }{ 10 }\) = \(\frac { 3 }{ 5 }\)
Question 3.
If sin A = \(\frac { 3 }{ 5 }\), prove that tan A + \(\frac { 1 }{ cos A }\) = 2 or – 2.
Solution:
Given sin A = \(\frac { 3 }{ 5 }\)
∴ cos A = \(\pm \sqrt{1-\sin ^2 A}= \pm \sqrt{1-\left(\frac{3}{5}\right)^2}\)
= \(\pm \sqrt{1-\frac{9}{25}}= \pm \frac{4}{5}\)
∴ tan A = \(\frac{\sin A}{\cos A}=\frac{3 / 5}{ \pm 4 / 5}= \pm \frac{3}{4}\)
Case-I : When tan A = \(\frac { 3 }{ 4 }\); cos A = \(\frac { 4 }{ 5 }\)
∴ tan A + \(\frac{1}{\cos A}=\frac{3}{4}+\frac{5}{4}=\frac{8}{4}\) = 2
Case-II : When tan A = – \(\frac { 3 }{ 4 }\); cos A = – \(\frac { 4 }{ 5 }\)
∴ tan A + \(\frac{1}{\cos A}=-\frac{3}{4}-\frac{5}{4}=\frac{-3-5}{4}\) = – 2
Thus, tan A + \(\frac { 1 }{ cos A }\) = 2 or – 2
Question 4.
If tan θ = \(\frac{1}{\sqrt{7}}\), find the value of \(\frac{{cosec}^2 \theta-\sec ^2 \theta}{{cosec}^2 \theta+\sec ^2 \theta}\).
Solution:
Question 5.
If sin θ = \(\frac { 21 }{ 29 }\), prove that sec θ + tan θ = 2\(\frac { 1 }{ 2 }\)
if θ lies between 0 and \(\frac { π }{ 2 }\). What will be the value of the expression when θ lies
(i) between \(\frac { π }{ 2 }\) and π and
(ii) between π and \(\frac { 3π }{ 2 }\) ?
Solution:
Given sin θ = \(\frac { 21 }{ 29 }\)
Since θ lies between 0 and \(\frac { π }{ 2 }\) ∴ cos θ > 0
∴ cos θ = + \(\sqrt{1-\sin ^2 \theta}\)
= \(\sqrt{1-\left(\frac{21}{29}\right)^2}=\sqrt{1-\frac{441}{841}}=\sqrt{\frac{400}{841}}\)
⇒ cos θ = \(\frac { 20 }{ 29 }\)
∴ tan θ = \(\frac{\sin \theta}{\cos \theta}=\frac{\frac{21}{29}}{\frac{20}{29}}=\frac{21}{20}\)
∴ sec θ + tan θ = \(\frac{29}{20}+\frac{21}{20}=\frac{50}{20}\)
= \(\frac{5}{2}=2 \frac{1}{2}\)
(i) When θ lies between \(\frac { π }{ 2 }\) and π
∴ cos θ, tan θ < 0
∴ cos θ = – \(\frac { 20 }{ 29 }\) and tan θ = \(\frac{\sin \theta}{\cos \theta}=-\frac{21}{20}\)
Thus sec θ + tan θ = – \(\frac{29}{20}-\frac{21}{20}\)
= –\(\frac{50}{20}=-\frac{5}{2}\)
(ii) When θ lies in π and \(\frac { 3π }{ 2 }\)
then sin θ < 0 but here sin θ = \(\frac { 21 }{ 29 }\) > 0
∴ (ii) is impossible.
Question 6.
If θ lies in the second quadrant and tan θ = \(\frac { -5 }{ 12 }\), find the value of \(\frac{2 \cos \theta}{1-\sin \theta}\).
Solution:
When θ lies in second quadrant then
sin θ > 0 and cos θ < 0
Question 7.
If sin θ sec θ = – 1 and θ lies in the second quadrant, find sin θ and sec θ.
Solution:
Given sin θ sec θ = – 1
⇒ tan θ = – 1
and θ lies in second quadrant
∴ sin θ > 0
while sec θ < 0.
Question 8.
If A is in the fourth quadrant and cos A = \(\frac { 5 }{ 13 }\), find the value of \(\frac{13 \sin A+5 \sec A}{5 \tan A+6 {cosec} A}\).
Solution:
Given cos A = \(\frac { 5 }{ 13 }\) and A lies in fourth quadrant.
Verify that
Question 9.
sin 60° = \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\).
Solution:
Question 10.
cos 60° = \(\frac{1-\tan ^2 30^{\circ}}{1+\tan ^2 30^{\circ}}\).
Solution:
Prove that
Question 11.
sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60° = \(\frac { 7 }{ 2 }\).
Solution:
L.H.S = sec 30° tan 60° 4- sin 45° cosec 45° + cos 30° cot 60°
= \(\left(\frac{2}{\sqrt{3}}\right) \sqrt{3}+\frac{1}{\sqrt{2}} \times \sqrt{2}+\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}=2+1+\frac{1}{2}=\frac{7}{2}\) = R.H.S
Question 12.
\(\frac { 4 }{ 3 }\) cot² 30° + 3 sin² 60° – 2 cosec 60° – \(\frac { 3 }{ 4 }\) tan² 30° = 3\(\frac { 1 }{ 3 }\).
Solution:
L.H.S = \(\frac { 4 }{ 3 }\) cot² 30° + 3 sin² 60° – 2 cosec 60° – \(\frac { 3 }{ 4 }\) tan² 30°
= \(\frac{4}{3}(\sqrt{3})^2+3\left(\frac{\sqrt{3}}{2}\right)^2-2\left(\frac{2}{\sqrt{3}}\right)^2-\frac{3}{4} \times\left(\frac{1}{\sqrt{3}}\right)^2=\frac{4}{3} \times 3+3 \times \frac{3}{4}-2 \times \frac{4}{3}-\frac{3}{4} \times \frac{1}{3}\)
= 4 + \(\frac{9}{4}-\frac{8}{3}-\frac{1}{4}=\frac{48+27-32-3}{12}=\frac{40}{12}=\frac{10}{3}=3 \frac{1}{3}\) = R.H.S