Accessing S Chand Class 11 Maths Solutions Chapter 4 Trigonometrical Functions Chapter Test can be a valuable tool for students seeking extra practice.
S Chand Class 11 ICSE Maths Solutions Chapter 4 Trigonometrical Functions Chapter Test
Question 1.
If cot A = \(\frac { 3 }{ 4 }\), find the value of 3 cos A + 5 sin A, where A lies in the first quadrant.
Solution:
Given cot A = \(\frac { 3 }{ 4 }\) ⇒ cosec A = ± \(\sqrt{1+\cot ^2 A}= \pm \sqrt{1+\left(\frac{3}{4}\right)^2}\)
⇒ cosec A = ± \(\sqrt{1+\frac{9}{16}}= \pm \frac{5}{4}\) ⇒ sin A = ± \(\frac { 4 }{ 5 }\)
But A lies in first quadrant
∴ sin A > 0, cos A > 0
∴ sin A = \(\frac { 4 }{ 5 }\)
Thus cos A = cot A sin A = \(\frac { 3 }{ 4 }\) x \(\frac { 4 }{ 5 }\) = \(\frac { 3 }{ 5 }\)
∴ 3 cos A + 5 sin A = \(3\left(\frac{3}{5}\right)+5\left(\frac{4}{5}\right)=\frac{9}{5}+4=\frac{29}{5}\)
Question 2.
If cos 120° = \(\frac { -1 }{ 2 }\) find the values of sin 120° and tan 120°.
Solution:
Given cos 120° = – \(\frac { 1 }{ 2 }\) since 120° lies in 2nd quadrant.
∴ sin 120° > 0 and tan 120° < 0
Now sin² θ + cos² θ = 1
⇒ sin 120° = + \(\sqrt{1-\cos ^2 120^{\circ}}=\sqrt{1-\left(-\frac{1}{2}\right)^2}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}\)
∴ tan 120° = \(\frac{\sin 120^{\circ}}{\cos 120^{\circ}}=\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=-\sqrt{3}\)
Question 3.
Prove that sec (- 1680°). sin 330° = 1.
Solution:
L.H.S = sec (- 1680°) sin 330° = sec (1680°) sin (360° – 30°) [∵ sec (- θ) = sec θ]
= sec (1440° + 240°) (- sin 30°) [∵ sin (360° – θ) = – sin θ]
= sec (180° + 60°) (- sin 30°) = (- sec 60°) (- sin 30°)
= (- 2)(-\(\frac { 1 }{ 2 }\)) = 1 = R.H.S
Question 4.
If A, B, C, D are the angles of a cyclic quadrilateral, show that cos A + cos B + cos C + cos D = 0
Solution:
Since A, B, C and D are the angles of a cyclic quadrilateral.
sum of opposite angles of cyclic quadrilateral by 180.
∴ A + C = B + D = 180° … (1)
L.H.S = cos A + cos B + cos C + cos D
= cos A + cos B + cos (180° – A) + cos (180° – B) [using eqn. (1)]
= cos A + cos B – cos A – cos B = 0
= R.H.S.
Question 5.
If tan 25° = a, prove that \(\frac{\tan 155^{\circ}-\tan 115^{\circ}}{1+\tan 115^{\circ} \cdot \tan 115^{\circ}}=\frac{1-a^2}{2 a}\).
Solution:
Question 6.
If A, B, C be the angles of a triangle, prove that
\(\frac{\sin (B+C)+\sin (C+A)+\sin (A+B)}{\sin (\pi+A)+\sin (3 \pi+B)+\sin (5 \pi+C)}\) = 1
Solution:
Since A, B, C are angles of triangle
∴ A + B + C = 180°
Question 7.
Prove that \(\frac{\cos \theta}{1-\sin \theta}+\frac{1-\sin \theta}{\cos \theta}\) = 2 sec θ.
Solution:
Question 8.
If sec θ = \(\sqrt{2}\) and \(\frac { 3π }{ 2 }\) < θ < 2π, find the value \(\frac{1+\tan \theta+{cosec} \theta}{1+\cot \theta-{cosec} \theta}\).
Solution:
