Summary
A galaxy is a huge assembly of stars, dust and gases, mutually held together by gravitational force . Millions of stars are present in the galaxy and the diameter of galaxy ranges from 1000 light years to one million light years. This note contains information on types and examples of galaxies and constellations.
Subjective Questions
Q1:
Without using calculator or table, find the value of cos75°.
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>cos75°</p> <p>= cos (45° + 30°)</p> <p>= cos45° ⋅ cos30° - sin45°⋅ sin30°</p> <p>= \(\frac {1}{\sqrt 2}\) ⋅\(\frac {\sqrt 3}{2}\) - \(\frac 1{\sqrt 2}\) ⋅\(\frac 12\)</p> <p>= \(\frac {\sqrt 3}{2\sqrt 2}\) - \(\frac 1{2\sqrt 2}\)</p> <p>= \(\frac {\sqrt 3 - 1}{2\sqrt 2}\) <sub>Ans</sub></p>
Q2:
Find the value of tan15° without using calculator or table.
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>tan15°</p> <p>= tan (60° - 45°)</p> <p>= \(\frac {tan60° - tan45°}{1 + tan60° ⋅tan45°}\)</p> <p>= \(\frac {\sqrt 3 - 1}{1 + \sqrt 3 ⋅ 1}\)</p> <p>= \(\frac {\sqrt 3 - 1}{\sqrt 3 + 1}\)× \(\frac {\sqrt 3 - 1}{\sqrt 3 - 1}\)</p> <p>= \(\frac {(\sqrt 3 - 1)^2}{{(\sqrt 3)^2}-{1^2}}\)</p> <p>= \(\frac {3 - 2 {\sqrt 3 + 1}}{3 - 1}\)</p> <p>= \(\frac {4 -2{\sqrt 3}}{2}\)</p> <p>= \(\frac {2(2 - \sqrt 3)}{2}\)</p> <p>= 2 - \(\sqrt 3\)<sub>Ans</sub></p>
Q3:
Find the value of sin 75° sin 15° without using calculator or table.
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>sin 75° sin 15°</p> <p>= sin (45° + 30°) sin (45° - 30°)</p> <p>= (sin 45° cos 30° + cos45° sin30°) (sin 45° cos 30° - cos 45° sin30°)</p> <p>= (\(\frac 1{\sqrt 2}\)⋅\(\frac {\sqrt 3}{2}\) + \(\frac 1{\sqrt 2}\)⋅\(\frac {1}{2}\))(\(\frac 1{\sqrt 2}\)⋅\(\frac {\sqrt 3}{2}\) -\(\frac 1{\sqrt 2}\)⋅\(\frac {1}{2}\))</p> <p>= (\(\frac {\sqrt 3}{2\sqrt 2}\))<sup>2</sup> -(\(\frac {1}{2\sqrt 2}\))<sup>2</sup></p> <p>= \(\frac 38\) - \(\frac 18\)</p> <p>= \(\frac {3 - 1}{8}\)</p> <p>= \(\frac 28\)</p> <p>= \(\frac 14\) <sub>Ans</sub></p>
Q4:
Find the value of cos 105° cos 15° without using calculator or table.
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>cos 105° cos 15°</p> <p>= cos (60° + 45°) cos (60° - 45°)</p> <p>= (cos 60° cos 45° - sin 60° sin 45°) (cos 60° cos 45° + sin 60° sin 45°)</p> <p>= (cos 60° cos 45°)<sup>2</sup> - (sin 60° sin 45°)<sup>2</sup></p> <p>= (\(\frac 12\) × \(\frac {1}{\sqrt 2}\))<sup>2</sup> - (\(\frac {\sqrt 3}{2}\) × \(\frac {1}{\sqrt 2}\))<sup>2</sup></p> <p>= \(\frac 18\) - \(\frac 38\)</p> <p>= \(\frac {1 - 3}{8}\)</p> <p>= \(\frac {-2}{8}\)</p> <p>= \(\frac {-1}{4}\) <sub>Ans</sub></p> <p></p>
Q5:
Without usinga calculator or a table, calculate the value of cos 105°.
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>cos 105°</p> <p>= cos (60° + 45°)</p> <p>= cos 60° cos 45° - sin 60° sin 45°</p> <p>= \(\frac 12\)⋅ \(\frac 1{\sqrt 2}\) -\(\frac {\sqrt 3}2\)⋅ \(\frac 1{\sqrt 2}\)</p> <p>= \(\frac {1 - \sqrt 3}{2\sqrt 2}\) <sub>Ans</sub></p>
Q6:
Prove that:
1 - tan 35° tan 10° = tan 35° + tan 10°
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>10° + 35° = 45°</p> <p>Putting tan on both sides,</p> <p>tan (10° + 35°) = tan 45°</p> <p>or, \(\frac {tan 10° + tan 35°}{1 - tan 10° tan 35°}\) = 1</p> <p>or, tan 10° + tan 35° = 1 - tan 10° tan35°</p> <p>or, 1 - tan 10° tan 35° = tan 10° + tan 35°</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q7:
If A + B = 45° , prove that:
(1 + tan A) (1 + tan B) = 2
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>A + B = 45°</p> <p>Putting tan on both;</p> <p>tan (A + B) = tan 45°</p> <p>or, \(\frac {tan A + tan B}{1 - tan A tan B}\) = 1</p> <p>or, tan A + tan B = 1 - tan A tan B</p> <p>or, tan A + tan B + tan A tan B = 1</p> <p>or, tan A + tan B + tan A tan B + 1 = 1 + 1</p> <p>or, tan A + tan A tan B + 1 + tan B = 2</p> <p>or, tan A (1 + tan B) + 1 (1 + tan B) = 2</p> <p>or, (1 + tan B) (tan A + 1) = 2</p> <p>∴ (1 + tan A) (1 + tan B) = 2</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q8:
Prove that:
cot (A - B) = \(\frac {cot A cot B + 1}{cot B - cot A}\)
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>L.H.S.</p> <p>= cot (A - B)</p> <p>= \(\frac {cos (A - B)}{sin (A - B)}\)</p> <p>= \(\frac {cos A cos B + sin A sin B}{sin A cos B - cos A sin B}\)</p> <p>=\(\frac {\frac {cos A cos B}{sin A sin B}+ \frac{sin A sin B}{sin A sin B}}{\frac {sin A cos B}{sin A sin B}+ \frac{cos A sin B}{sin A sin B}}\)</p> <p>= \(\frac {cot A cot B + 1}{cot B - cot A}\)</p> <p>= R.H.S <sub>Proved</sub></p>
Q9:
If A + B = 45°, show that:
tan A + tan B + tan A⋅tan B = 1
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>A + B = 45°</p> <p>Taking tan on both sides:</p> <p>tan (A + B) = tan 45°</p> <p>or, \(\frac {tan A + tan B}{1 - tan A tan B}\) = 1</p> <p>or, tan A + tan B = 1 - tan A tan B</p> <p>or, tan A + tan B + tan A tan B = 1</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q10:
Prove that:
1 - tan 20° tan 25° = tan 20° + tan 25°
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>20° + 25° = 45°</p> <p>Taking tan on both sides,</p> <p>tan (20° + 25°) = tan 45°</p> <p>or, \(\frac {tan 20° + tan 25°}{1 - tan 20° tan 25°}\) = 1</p> <p>or, tan 20° + tan 25° = 1 - tan 20° tan 25°</p> <p>∴ 1 - tan 20° tan 25° = tan 20° + tan 25°</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q11:
If tan α = \(\frac 56\) and tan β = \(\frac 1{11}\), prove that:
α + β = \(\frac {π^2}4\)
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>tan (α + β) = \(\frac {tanα + tanβ}{1 - tan α tan β}\)</p> <p>or, tan (α + β) =\(\frac {\frac 56 + \frac 1{11}}{1 - \frac56 × \frac {1}{11}}\)</p> <p>or, tan (α + β) =\(\cfrac {\frac {55 + 5}{66}}{\frac {66 - 5}{66}}\)</p> <p>or, tan (α + β) =\(\cfrac {\frac {61}{66}}{\frac {61}{66}}\)</p> <p>or, tan (α + β) = \(\frac {61}{66}\)× \(\frac {66}{61}\)</p> <p>or, tan (α + β) = 1</p> <p>or, tan (α + β) = tan \(\frac {π^c}{4}\)</p> <p>∴ (α + β) = \(\frac {π^c}{4}\) <sub>Proved</sub></p>
Q12:
Without using table, prove that:
sin 105° + cos 105° = \(\frac {1}{\sqrt 2}\)
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>L.H.S.</p> <p>= sin 105° + cos 105°</p> <p>= sin (60° + 45°) + cos (60° + 45°)</p> <p>= sin 60° cos 45° + cos 60° sin 45° + cos 60° cos 45° - sin 60° sin 45°</p> <p>= \(\frac {\sqrt3}{2}\)⋅\(\frac 1{\sqrt 2}\) + \(\frac 12\)⋅\(\frac 1{\sqrt 2}\) + \(\frac 12\)⋅\(\frac 1{\sqrt 2}\) - \(\frac {\sqrt 3}2\)⋅\(\frac 1{\sqrt 2}\)</p> <p>= \(\frac {\sqrt 3}{2\sqrt 2}\) + \(\frac 1{2\sqrt 2}\) + \(\frac 1{2\sqrt 2}\) - \(\frac {\sqrt 3}{2\sqrt 2}\)</p> <p>= \(\frac {1 + 1}{2\sqrt 2}\)</p> <p>= \(\frac 2{2\sqrt 2}\)</p> <p>= \(\frac 1{\sqrt 2}\)</p> <p>= R.H.S.</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q13:
If m sin (α + θ) = n sin (β + θ), prove that :
cot θ = \(\frac {m cos α - n cos β}{n sin β - m sin α}\)
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>Given,</p> <p>m sin(α + θ) = n sin (β + θ)</p> <p>or, m (sin α cos θ + cos α sin θ) = n (sin β cos θ + cos β sin θ)</p> <p>or, msin α cos θ + mcos α sin θ = nsin β cos θ + ncos β sin θ</p> <p>or, msin α cos θ -nsin β cos θ =ncos β sin θ -mcos α sin θ</p> <p>or,cos θ (msin α - nsin β) = sin θ (ncos β - mcos α)</p> <p>or, \(\frac {cos θ}{sin θ}\) = \(\frac {ncos β - mcos α}{msin α - nsin β}\)</p> <p>∴ cot θ = \(\frac {ncos β - mcos α}{msin α - nsin β}\) <sub>Proved </sub></p>
Q14:
Prove that:
tan 20° + tan 72° + tan 88° = tan 20° ⋅ tan 72° ⋅ tan 88°
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>20° + 72° + 88° = 180°</p> <p>20° + 72° = 180° - 88°</p> <p>Putting tan on both,</p> <p>tan (20° + 72°) = tan (180° - 88°)</p> <p>or, \(\frac {tan 20° + tan 72°}{1 - tan 20° tan 72°}\) = 0 - tan 88°</p> <p>or, tan 20° + tan 72° = - tan 88° (1 - tan 20° tan 72°)</p> <p>or, tan 20° + tan 72° = - tan 88° + tan 20° tan 72° tan 88°</p> <p>or, tan 20° + tan 72° + tan 88° = tan 20° tan 72° tan 88°</p> <p>Hence L.H.S. = R.H.S. <sub>Proved</sub></p>
Q15:
Prove that:
sin (x + y) - sin (x - y) = 2cos x siny
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>L.H.S.</p> <p>= sin (x + y) - sin ( x - y)</p> <p>= sin x cos y + cos x sin y - (sin x cos y - cos x sin y)</p> <p>=sin x cos y + cos x sin y - sin x cos y + cos x sin y</p> <p>= 2 cos x sin y</p> <p>= R.H.S.</p> <p>Hence L.H.S. = R.H.S. <sub>Proved</sub></p>
Q16:
If A + B = \(\frac {π^c}{4}\), prove that:
(cot A - 1) (cot B - 1) = 2
Type: Short
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>A + B = \(\frac {π^c}{4}\)</p> <p>Taking cot on both,</p> <p>cot (A + B) = cot \(\frac {π^c}{4}\)</p> <p>or, \(\frac {cot A cot B - 1}{cot B + cot A}\) = 1</p> <p>or, cot A cot B - 1 = cot B + cot A</p> <p>or, cot A cot B - cot A - cot B = 1</p> <p>or, cot A cot B - cot A - cot B + 1 = 1 + 1</p> <p>or, cot A (cot B - 1) -1 (cot B - 1) = 2</p> <p>or, (cot A - 1) (cot B - 1) = 2</p> <p>∴ L.H.S. = R.H.S. <sub>Proved</sub></p> <p></p>
Q17:
Without using calculator or a table, prove that:
tan 55° - tan 35° = 2 tan 20°
Type: Long
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>55° - 35° = 20°</p> <p>Taking tan on both sides,</p> <p>tan (55° - 35°) = tan 20°</p> <p>or, \(\frac {tan 55° - tan 35°}{1 + tan 55° tan35°}\) = tan 20°</p> <p>or, tan 55° - tan 35° = tan 20° (1 + tan 55° tan35°)</p> <p>or,tan 55° - tan 35° = tan 20° + tan 20° tan35° tan 55°</p> <p>or, tan 55° - tan 35° = tan 20° + tan 20° tan (90° - 55°) tan 55°</p> <p>or, tan 55° - tan 35° = tan 20° + tan 20° cot 55° tan 55°</p> <p>or, tan 55° - tan 35° = tan 20° + tan 20°</p> <p>∴ tan 55° - tan 35° = 2 tan 20°</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q18:
Prove that:
cos (A+ B + C) = cos A cos B cos C (1 - tan B tan C - tan C tan B - tan A tan B)
Type: Long
Difficulty: Easy
Show/Hide Answer
Answer: <p>L.H.S.</p> <p>= cos (A + B + C)</p> <p>= cos (A + B) cos C - sin (A + B) sin C</p> <p>= (cos A cos B - sin A sin B) cos C - (sin A cos B + cos A sin B) sin C</p> <p>multiply by \(\frac {cos A cos B cos C}{cos A cos B cos C}</p> <p>= cos A cos B cos C (\(\frac {cos A cos B cos C}{cos A cos B cos C}\) - \(\frac {sin A sin B sin C}{cos A cos B cos C}\) - \(\frac {sin A cos Bsin C}{cos A cosB cos C}\) - \(\frac {cos A sin B sin C}{cos A cos B cos C}\))</p> <p>= cos A cos B cos C (1 - tan A tan B - tan A tan C - tan B tan C)</p> <p>= cos A cos B cos C (1 - tan B tan C - tan C tan A - tan A tan B)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q19:
Prove that:
\(\frac {cos 17° + sin 17°}{cos 17° - sin 17°}\) = tan 62°
Type: Long
Difficulty: Easy
Show/Hide Answer
Answer: <p>R.H.S.</p> <p>= tan 62°</p> <p>= \(\frac {sin 62°}{cos 62°}\)</p> <p>= \(\frac {sin (45° + 17°)}{cos (45° + 17°)}\)</p> <p>= \(\frac {sin 45° cos 17° + cos 45° sin 17°}{cos 45° cos 17° - sin 45° sin 17°}\)</p> <p>= \(\frac {\frac {1}{\sqrt 2} cos 17° + \frac {1}{\sqrt 2} sin 17°}{\frac {1}{\sqrt 2} cos 17°- \frac {1}{\sqrt 2} sin 17°}\)</p> <p>= \(\frac {\frac {1}{\sqrt 2} (cos 17° + sin 17°)}{\frac {1}{\sqrt 2} (cos 17°- sin 17°)}\)</p> <p>= \(\frac {cos 17° + sin 17°}{cos 17° - sin 17°}\)</p> <p>= L.H.S. <sub>Proved</sub></p>
Q20:
Prove that:
\(\frac {cos 8° + sin 8°}{cos 8° - sin 8°}\) = tan 53°
Type: Long
Difficulty: Easy
Show/Hide Answer
Answer: <p>R.H.S.</p> <p>= tan 53°</p> <p>= \(\frac {sin 53°}{cos 53°}\)</p> <p>= \(\frac {sin (45° + 8°)}{cos (45° + 8°)}\)</p> <p>= \(\frac {sin 45° cos 8° + cos 45° sin 8°}{cos 45° cos 8° - sin 45° sin 8°}\)</p> <p>= \(\frac {\frac {1}{\sqrt 2} cos 8° + \frac {1}{\sqrt 2} sin 8°}{\frac {1}{\sqrt 2} cos 8°- \frac {1}{\sqrt 2} sin 8°}\)</p> <p>= \(\frac {\frac {1}{\sqrt 2} (cos 8° + sin 8°)}{\frac {1}{\sqrt 2} (cos 8°- sin 8°)}\)</p> <p>= \(\frac {cos 8° + sin 8°}{cos 8° - sin 8°}\)</p> <p>= L.H.S. <sub>Proved</sub></p>
Q21:
If sin A = \(\frac 35\) and cos B = \(\frac 5{13}\), find the value of cos (A +B).
Type: Long
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>sin A = \(\frac 35\) and cos B = \(\frac 5{13}\)</p> <p>cos A = \(\sqrt {1 - sin^2 A}\)</p> <p>= \(\sqrt {1 - (\frac 35)^2}\)</p> <p>= \(\sqrt {1 - \frac 9{25}}\)</p> <p>= \(\sqrt {\frac {25 - 9}{25}}\)</p> <p>= \(\sqrt {\frac {16}{25}}\)</p> <p>= \(\frac 45\)</p> <p>sin B = \(\sqrt {1 - cos^2 B}\)</p> <p>= \(\sqrt {1 - (\frac 5{13})^2}\)</p> <p>= \(\sqrt {1 - \frac {25}{169}}\)</p> <p>= \(\sqrt {\frac {169 - 25}{169}}\)</p> <p>= \(\sqrt {\frac {144}{169}}\)</p> <p>= \(\frac {12}{13}\)</p> <p>Now,</p> <p>cos (A + B) = cos A cos B - sin A sin B</p> <p>= \(\frac 45\)⋅ \(\frac 5{13}\) - \(\frac 35\)⋅ \(\frac {12}{13}\)</p> <p>= \(\frac {20}{65}\) - \(\frac {36}{65}\)</p> <p>= \(\frac {20 - 36}{65}\)</p> <p>= -\(\frac {16}{65}\) <sub>Ans</sub></p>
Q22:
If sin A = \(\frac 1{\sqrt {10}}\), sin B = \(\frac 1{\sqrt 5}\), prove that:
A + B = \(\frac {π^2}{4}\)
Type: Long
Difficulty: Easy
Show/Hide Answer
Answer: <p>Here,</p> <p>sin A = \(\frac 1{\sqrt {10}}\), sin B = \(\frac 1{\sqrt 5}\)</p> <p>cos A = \(\sqrt {1 - sin^2 A}\)</p> <p>= \(\sqrt {1 - (\frac 1{\sqrt {10}})^2}\)</p> <p>= \(\sqrt {1 - \frac 1{10}}\)</p> <p>= \(\sqrt {\frac {10 - 1}{10}}\)</p> <p>= \(\sqrt {\frac {9}{10}}\)</p> <p>= \(\frac 3{\sqrt {10}}\)</p> <p>cos B = \(\sqrt {1 - sin^2 B}\)</p> <p>= \(\sqrt {1 - (\frac 1{\sqrt 5})^2}\)</p> <p>= \(\sqrt {1 - \frac 15}\)</p> <p>= \(\sqrt {\frac {5 - 1}{5}}\)</p> <p>= \(\sqrt {\frac {4}{5}}\)</p> <p>= \(\frac 2{\sqrt 5}\)</p> <p>sin (A + B) = sin A cos B + cos A sin B</p> <p>or, sin (A + B) = \(\frac 1{\sqrt {10}}\)× \(\frac 2{\sqrt 5}\) + \(\frac 3{\sqrt {10}}\)× \(\frac 1{\sqrt 5}\)</p> <p>or, sin (A + B) = \(\frac 2{\sqrt {50}}\) + \(\frac 3{\sqrt {50}}\)</p> <p>or, sin (A + B) = \(\frac {2 + 3}{\sqrt {50}}\)</p> <p>or, sin (A + B) = \(\frac 5{\sqrt {50}}\)</p> <p>or, sin (A + B) = \(\frac 5{5\sqrt 2}\)</p> <p>or, sin (A + B) = \(\frac 1{\sqrt 2}\)</p> <p>or, sin (A + B) = sin \(\frac {π^2}4\)</p> <p>∴ A + B =\(\frac {π^2}4\) <sub>Proved</sub></p>
Q23:
Prove that:
\(\frac {cos 35° + sin 35°}{cos 35° - sin 35°}\) = cot 10°
Type: Long
Difficulty: Easy
Show/Hide Answer
Answer: <p>L.H.S.</p> <p>= \(\frac {cos 35° + sin 35°}{cos 35° - sin 35°}\)</p> <p>=\(\frac {cos (45° - 10°) + sin (45° - 10°)}{cos (45° - 10°) - sin (45° - 10°)}\)</p> <p>= \(\frac {cos 45° cos 10° + sin 45° sin 10° + sin 45° cos 10° - cos 45° sin 10°}{cos 45° cos 10° + sin 45° sin 10° - sin 45° cos 10° + cos 45° sin 10°}\)</p> <p>= \(\frac {\frac 1{\sqrt 2} cos 10° + \frac 1{\sqrt 2} sin 10° + \frac 1{\sqrt 2} cos 10° - \frac 1{\sqrt 2} sin 10°}{\frac 1{\sqrt 2} cos 10° + \frac 1{\sqrt 2} sin 10° - \frac 1{\sqrt 2} cos 10° + \frac 1{\sqrt 2} sin 10°}\)</p> <p>=\(\frac {\frac 2{\sqrt 2} cos 10°}{\frac 2{\sqrt 2} sin 10°}\)</p> <p>= \(\frac {cos 10°}{sin 10°}\)</p> <p>= cot 10°</p> <p>= R.H.S.<sub>Proved</sub></p> <p></p>
Q24:
Prove that:
cos (A + B)⋅cos (A - B) = cos2A - sin2B = cos2B - sin2A
Type: Long
Difficulty: Easy
Show/Hide Answer
Answer: <p>L.H.S.</p> <p>=cos (A + B)⋅cos (A - B)</p> <p>= (cos A cos B - sin a sin B)(cos A cos B + sin A sin B)</p> <p>= (cos A cos B)<sup>2</sup> - (sin A sin B)<sup>2</sup></p> <p>= cos<sup>2</sup>A cos<sup>2</sup>B - sin<sup>2</sup>A sin<sup>2</sup>B</p> <p>= cos<sup>2</sup>A (1 - sin<sup>2</sup>B) - (1 - cos<sup>2</sup>A) sin<sup>2</sup>B</p> <p>= cos<sup>2</sup>A - cos<sup>2</sup>A sin<sup>2</sup>B - sin<sup>2</sup>B + cos<sup>2</sup>A sin<sup>2</sup>B</p> <p>= cos<sup>2</sup>A - sin<sup>2</sup>B (M.H.S.)</p> <p>Again,</p> <p>cos<sup>2</sup>A - sin<sup>2</sup>B</p> <p>= (1 - sin<sup>2</sup>A) - (1 - cos<sup>2</sup>B)</p> <p>= 1 - sin<sup>2</sup>A - 1 + cos<sup>2</sup>B</p> <p>= cos<sup>2</sup>B - sin<sup>2</sup>A</p> <p>∴ L.H.S. = M.H.S. = R.H.S. <sub>Proved</sub></p>
Q25:
An angle θ is divided into two parts α and β such that tanα:tanβ = x:y show that:
sin(α - β) = \(\frac {x - y}{x + y}\) sinθ
Type: Long
Difficulty: Easy
Show/Hide Answer
Answer: <p>Given,</p> <p>θ =α +β</p> <p>sinθ = sin(α + β)............................(1)</p> <p>And</p> <p>\(\frac {tanα}{tanβ}\) = \(\frac xy\)</p> <p>\(\frac {tanα - tanβ}{tanα + tanβ}\) = \(\frac {x - y}{x + y}\).............................(2)</p> <p>Now,</p> <p>R.H.S.</p> <p>= \(\frac {x - y}{x + y}\) sinθ</p> <p>=\(\frac {tanα - tanβ}{tanα + tanβ}\) sinθ</p> <p>=\(\frac {\frac {sinα}{cosα} -\frac {sinβ}{cosβ}}{\frac {sinα}{cosα} + \frac {sinβ}{cosβ}}\) sinθ</p> <p>= \(\frac{\frac{sin\alpha.cos\beta - cos\alpha.sin\beta}{cos\alpha.cos\beta}}{\frac{sin\alpha.cos\beta + cos\alpha.sin\beta}{cos\alpha.cos\beta}}\). sin\(\theta\)</p> <p>= \(\frac {sin\alpha cos\beta - cos\alpha sin\beta}{sin\alpha cos\beta + cos\alpha sin\beta}\) . sin\(\theta\)</p> <p>= \(\frac {sin (\alpha - \beta)}{sin(\alpha + \beta)}\) . sin\(\theta\)</p> <p>= \(\frac {sin(\alpha - \beta)}{sin \theta}\). sin\(\theta\)</p> <p>= sin (α - β)</p> <p>Hence, L.H.S. = R.H.S. <sub>Proved</sub></p>
Q26:
Prove that:
sin(A + B + C) = cosA . cosB . cosC(tanA + tanB + tanC - tanA . tanB . tanC)
Type: Long
Difficulty: Easy
Show/Hide Answer
Answer: <p>L.H.S.</p> <p>= sin (A + B + C)</p> <p>= sin {A + (B + C)}</p> <p>= sinA . cos(B + C) + cos A . cos (B + C)</p> <p>= sinA {cosB cosC - sinB sinC} + cosA {sinB cosC + cosB sinC}</p> <p>= sinA cosB cosC - sinA sinB sinC + cosA sinB cosC + cosA cosB sinC</p> <p>= cosA cosB cosC[\(\frac {sinA cosB cosC - sinA sinB sinC + cosA sinB cosC + cosA cosB sinC}{cosA cosB cosC}\)]</p> <p>= cosA cosB cosC [tanA - tanA tanB tanC + tanB + tanC]</p> <p>= cosA cosB cosC [tanA + tanB + tanC- tanA tanB tanC ]</p> <p>∴ L.H.S. = R.H.S. <sub>Proved</sub></p>
Q27: Cos36°=?5+1/4
Type: Long
Difficulty: Easy
Q28: IF 1/SinA+1/CosA=1/SinB+1/CosB ,PROVE THAT: Cot{(A+B)/2}=TanA.TanB
Type: Long
Difficulty: Easy
Q29: Prove that: cot (A - B) = cotAcotB+1cotB?cotA
Type: Short
Difficulty: Easy
Elliptical galaxy:
Spiral galaxy:
Irregular galaxy:
