M l aggarwal understanding icse mathematics class 9 Solutions for Chapter 18 trigonometric ratios and standard angles

Find the values of:

(i) sin2 60° – cos2 45° + 3tan2 30°

\begin{aligned} &\text { (ii) }\\ &\frac{2 \cos ^{2} 45^{\circ}+3 \tan ^{2} 30^{\circ}}{\sqrt{3} \cos 30^{\circ}+\sin 30^{\circ}} \end{aligned}

(iii) sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60°

Taking A = 30°, verify that

(i) cos4 A – sin4 A = cos 2A

(ii) 4cos A cos (60° – A) cos (60° + A) = cos 3 A.

If A = 45° and B = 30°, verify that sin A/ (cos A + sin A + sin B) = 2/3

Taking A = 60° and B = 30°, verify that

(i) sin (A + B)/ cos A cos B = tan A + tan B

(ii) sin (A – B)/ sin A sin B = cot B – cot A

\text { If } \sqrt{2} \tan 2 \theta=\sqrt{6} \text { and } \theta^{\circ}<2 \theta<90^{\circ}, \text { find the value of } \sin \theta+\sqrt{3} \cos \theta-2 \tan ^{2} \theta

If 3θ is an acute angle, solve the following equations for θ:

(i) (cosec 3θ – 2) (cot 2θ – 1) = 0

(ii) (tan θ – 1) (cosec 3θ – 1) = 0

If tan (A + B) = √3 and tan (A – B) = 1 and A, B (B < A) are acute angles, find the values of A

and B.

Without using trigonometrical tables, evaluate the following:

(i) sin2 28° + sin2 62° – tan2 45°

\begin{aligned} &\text { (ii) }\\ &2 \frac{\cos 27^{\circ}}{\sin 63^{\circ}}+\frac{\tan 27^{\circ}}{\cot 63^{\circ}}+\cos 0^{\circ} \end{aligned}

(iii) cos 18° sin 72° + sin 18° cos 72°

(iv) 5 sin 50° sec 40° – 3 cos 59° cosec 31°

Prove that:

\frac{\cos \left(90^{\circ}-\theta\right) \sec \left(90^{\circ}-\theta\right) \tan \theta}{\operatorname{cosec}\left(90^{\circ}-\theta\right) \sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right)}+\frac{\tan \left(90^{\circ}-\theta\right)}{\cot \theta}=2

When 0° < A < 90°, solve the following equations :

(i) sin 3A = cos 2A

(ii) tan 5A = cotA

Find the value of θ if

(i) sin (θ + 36°) = cos θ, where θ and θ + 36° are acute angles.

(ii) sec 4θ = cosec (θ – 20°), where 4θ and θ – 20° are acute angles.

In the adjoining figure, ABC is a right-angled triangle at B and ABD is right-angled triangle at

A. If BD ⊥ AC and BC = 2cm, find the length of AD.