Question 1 iEvaluate the following-i sin 60- cos 30- - sin 30- cos 60-

Name: NCERT - Grade 10 - Mathematics - Introduction to Trigonometry - Q12_1
Uploaded: 2022-07-04T10:36:42+05:30
Description: (i) sin 60^∘ cos 30^∘ + sin 30^∘ cos 60^∘ = (√(3)/2×√(3)/2)+ (1/2×1/2) = 3/4+ 1/4 = 4/4= 1 ...

(i) sin 60^∘ cos 30^∘ + sin 30^∘ cos 60^∘ = (√(3)/2×√(3)/2)+ (1/2×1/2) = 3/4+ 1/4 = 4/4= 1 ...

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Standard X

Mathematics

Trigonometric Ratios of Complementary Angles

Question 1 iE...

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Question 1 (i)
Evaluate the following:
(i) $s i n 60^{\circ} c o s 30^{\circ} + s i n 30^{\circ} c o s 60^{\circ}$

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Solution

(i) $s i n 60^{\circ} c o s 30^{\circ} + s i n 30^{\circ} c o s 60^{\circ}$
$= (\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}) + (\frac{1}{2} \times \frac{1}{2}) = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1$

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Q. Question 1 (i)
Evaluate the following:
(i)

s i n 60^{\circ} c o s 30^{\circ} + s i n 30^{\circ} c o s 60^{\circ}

Q. 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°

Q. Show that:
(i)

\frac{1 - \sin 60 °}{\cos 60 °} = \frac{\tan 60 ° - 1}{\tan 60 ° + 1}

(ii)

\frac{\cos 30 ° + \sin 60 °}{1 + \sin 30 ° + \cos 60 °} = \cos 30 °

Q. Find the value of the following

(i) sin 60^{\circ} cos 30^{\circ} + cos 60^{\circ} sin 30^{\circ}

(i i) cos 30^{\circ} cos 60^{\circ} - sin 30^{\circ} sin 60^{\circ}

Show that:

(i) $\frac{1 - sin 60^{\circ}}{cos 60^{\circ}} = \frac{tan 60^{\circ} - 1}{tan 60^{\circ} + 1}$

(ii) $\frac{cos 30^{\circ} + sin 60^{\circ}}{1 + sin 30^{\circ} + cos 60^{\circ}} = cos 30^{\circ}$

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Trigonometric Ratios of Complementary Angles

Standard X Mathematics

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Question 1 (i) Evaluate the following: (i) sin60∘cos30∘+sin30∘cos60∘

(i) sin60∘cos30∘+sin30∘cos60∘ =(√32×√32)+(12×12)=34+14=44=1

Question 1 (i)
Evaluate the following:
(i) $s i n 60^{\circ} c o s 30^{\circ} + s i n 30^{\circ} c o s 60^{\circ}$

(i) $s i n 60^{\circ} c o s 30^{\circ} + s i n 30^{\circ} c o s 60^{\circ}$
$= (\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}) + (\frac{1}{2} \times \frac{1}{2}) = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1$