Evaluate:
\sin 60^{\circ} \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ}
cos 60° cos 30° − sin 60° sin 30°.
cos 45° cos 30° + sin 45° sin 30°.
\frac{\sin 30^{\circ}}{\cos 45^{\circ}}+\frac{\cot 45^{\circ}}{\sec 60^{\circ}}-\frac{\sin 60^{\circ}}{\tan 45^{\circ}}+\frac{\cos 30^{\circ}}{\sin 90^{\circ}}
\left(5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}\right) /\left(\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}\right)
2 \cos ^{2} 60^{\circ}+3 \sin ^{2} 45^{\circ}-3 \sin ^{2} 30^{\circ}+2 \cos ^{2} 90^{\circ}
\cot ^{2} 30^{\circ}-2 \cos ^{2} 30^{\circ}-3 / 4 \sec ^{2} 45^{\circ}+1 / 4 \operatorname{cosec}^{2} 30^{\circ}
\left(\sin ^{2} 30^{\circ}+4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}\right)\left(\operatorname{cosec}^{2} 45^{\circ} \sec ^{2} 30^{\circ}\right)
4 / \cot ^{2} 30^{\circ}+1 / \sin ^{2} 30^{\circ}-2 \cos ^{2} 45^{\circ}-\sin ^{2} 0^{\circ}
Show that:
\text { (i) } \frac{1-\sin 60^{\circ}}{\cos 60^{\circ}}=\frac{\tan 60^{\circ}-1}{\tan 60^{\circ}+1}
\text { (ii) } \frac{\cos 30^{\circ}+\sin 60^{\circ}}{1+\sin 30^{\circ}+\cos 60^{\circ}}=\cos 30^{\circ}
Verify each of the following:
(i) sin 60° cos 30° − cos 60° sin 30° = sin 30°
(ii) cos 60° cos 30° + sin 60° sin 30° = cos 30°
(iii) 2 sin 30° cos 30° = sin 60°
(iv) 2 sin 45° cos 45° = sin 90°
If A = 45°, verify that:
(i) sin 2A = 2 sin A cos A
\text { (ii) } \cos 2 A=2 \cos ^{2} A-1=1-2 \sin ^{2} A
If A = 30°, verify that:
\text { (i) } \sin 2 A=\frac{2 \tan A}{1-\tan ^{2} A}
\text { (ii) } \cos 2 A=\frac{1-\tan ^{2} A}{1+\tan ^{2} A}
\text { (iii) } \tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}
If A = 60° and B = 30°, verify that:
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B − sin A sin B
(i) sin (A − B) = sin A cos B − cos A sin B
(ii) cos (A − B) = cos A cos B + sin A sin B
\text { (iii) } \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}
If A and B are acute angles such that tan A = 1/3, tan B = 1/2 and
\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}
Show that A + B = 45°.
Using the formula,
\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}
find the value of tan 60°, it being given that tan 30° = 1/√3.
\cos A=\sqrt{\frac{1+\cos 2 A}{2}}
find the value of cos 30°, it being given that cos 60° = 1/2.
\sin A=\sqrt{\frac{1-\cos 2 A}{2}}
find the value of sin 30°, it being given that cos 60° = 1/2.
In the adjoining figure, ∆ABC is a right-angled triangle in which ∠B = 90°, ∠A = 30° and AC = 20 cm.
Find (i) BC, (ii) AB.
In the adjoining figure, ∆ ABC is a right-angled at B and ∠A = 30°. If BC = 6 cm,
Find (i) AB, (ii) AC.
If sin (A + B) = 1 and cos (A − B) = 1, 0° ≤ (A + B) ≤ 90° and A > B, then find A and B.
If sin (A − B) = 1/2 and cos (A + B) = 1/2, 0° < (A + B) < 90° and A > B, then find A and B.
If tan (A − B) = 1/√3 and tan (A + B) = √3, 0° < (A + B) < 90° and A > B, then find A and B.
If 3x = cosec θ and 3/x = cot θ, find the value of 3(x^2 – 1/x^2).
If sin(A+B) = sinA cosB+cosA sinB and cos(A-B)=cosA cosB+sinA sinB,
find the values of (i) sin75° and (ii) cos15°.
Lido
Courses
Quick Links
Terms & Policies