Question 1: If \sin A = \frac{4}{5} and \cos B = \frac{5}{13} , where 0 < A , B < \frac{\pi}{2} , find the values of the following:   i) \sin (A+B)    ii) \cos (A+B)    iii) \sin (A - B)   iv) \cos(A-B)

Answer:

Given \sin A = \frac{4}{5} and \cos B = \frac{5}{13} , where 0 < A , B < \frac{\pi}{2}

This means that A is in Quadrant I and B is in Quadrant I

Therefore, \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \Big( \frac{4}{5} \Big)^2 } = \sqrt{\frac{9}{25}} = \frac{3}{5}

Similarly, \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \Big( \frac{5}{13} \Big)^2 } = \sqrt{\frac{144}{169}} = \frac{12}{13}

i) \sin ( A+B) = \sin A \cos B + \cos A \sin B = \frac{4}{5} \times \frac{5}{13} + \frac{3}{5} \times \frac{12}{13} = \frac{20+36}{65} = \frac{56}{65}

ii) \cos ( A+B) = \cos A \cos B - \sin A \sin B = \frac{3}{5} \times \frac{5}{13} - \frac{4}{5} \times \frac{12}{13} = \frac{15-48}{65} = \frac{-33}{65}

iii) \sin ( A-B) = \sin A \cos B - \cos A \sin B = \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13} = \frac{20-36}{65} = \frac{-16}{65}

iv) \cos ( A-B) = \cos A \cos B + \sin A \sin B = \frac{3}{5} \times \frac{5}{13} + \frac{4}{5} \times \frac{12}{13} = \frac{15+48}{65} = \frac{63}{65}

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Question 2:

a) If \sin A = \frac{12}{13} and \sin B = \frac{4}{5} , where \frac{\pi}{2} < A  < \pi , 0 < B < \frac{\pi}{2} , find the values of the following:   i) \sin (A+B)    ii) \cos (A+B)

b) If \sin A = \frac{3}{5} and \cos B = - \frac{12}{13} , where A and B both lie in Q II find the value of \sin (A+B)

Answer:

a) Given \sin A = \frac{12}{13} and \sin B = \frac{4}{5} , where \frac{\pi}{2} < A  < \pi , 0 < B < \frac{\pi}{2}

This means that A is in Quadrant II and B is in Quadrant I

Therefore, \cos A = -\sqrt{1 - \sin^2 A} = - \sqrt{1 - \Big( \frac{12}{13} \Big)^2 } = - \sqrt{\frac{25}{69}} = - \frac{5}{13}

Similarly, \cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \Big( \frac{4}{5} \Big)^2 } = \sqrt{\frac{9}{25}} = \frac{3}{5}

i) \sin ( A+B) = \sin A \cos B + \cos A \sin B = \frac{12}{13} \times \frac{3}{5} + \frac{-5}{13} \times \frac{4}{5} = \frac{36-20}{65} = \frac{-16}{65}

ii) \cos ( A+B) = \cos A \cos B - \sin A \sin B = \frac{-5}{13} \times \frac{3}{5} - \frac{12}{13} \times \frac{4}{5} = \frac{-15-48}{65} = \frac{-63}{65}

 

b) Given \sin A = \frac{3}{5} and \cos B = - \frac{12}{13} , where A and B both lie in Q II

Therefore, \cos A = -\sqrt{1 - \sin^2 A} = - \sqrt{1 - \Big( \frac{3}{5} \Big)^2 } = - \sqrt{\frac{16}{25}} = - \frac{4}{5}

Similarly, \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \Big( \frac{-12}{13} \Big)^2 } = \sqrt{\frac{25}{169}} = \frac{5}{13}

\sin ( A+B) = \sin A \cos B + \cos A \sin B = \frac{3}{5} \times \frac{-12}{13} + \frac{-4}{5} \times \frac{5}{13} = \frac{-36-20}{65} = \frac{-56}{65}

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Question 3: If \cos A = - \frac{24}{25} and \cos B = \frac{3}{5} , where \pi < A  < \frac{3\pi}{2} , \frac{3\pi}{2} < B < 2\pi , find the values of the following:   i) \sin (A+B)    ii) \cos (A+B)

Answer:

Given \cos A = - \frac{24}{25} and \cos B = \frac{3}{5} , where \pi < A  < \frac{3\pi}{2} , \frac{3\pi}{2} < B < 2\pi

This means that A is in Quadrant III and B is in Quadrant IV

Therefore, \sin A = -\sqrt{1 - \cos^2 A} = - \sqrt{1 - \Big( \frac{-24}{25} \Big)^2 } = - \sqrt{\frac{49}{625}} = - \frac{7}{25}

Similarly, \sin B = -\sqrt{1 - \cos^2 B} = - \sqrt{1 - \Big( \frac{3}{5} \Big)^2 } = - \sqrt{\frac{16}{25}} = - \frac{4}{5}

i) \sin ( A+B) = \sin A \cos B + \cos A \sin B = \frac{-7}{25} \times \frac{3}{5} + \frac{-24}{25} \times \frac{-4}{5} = \frac{-21+96}{125} = \frac{3}{5}

ii) \cos ( A+B) = \cos A \cos B - \sin A \sin B = \frac{-24}{25} \times \frac{3}{5} - \frac{-7}{25} \times \frac{-4}{5} = \frac{-72-28}{125} = \frac{-4}{5}

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Question 4: If \tan A = \frac{3}{4} and \cos B = \frac{9}{41} , where \pi < A  < \frac{3\pi}{2} and  0 < B < \frac{\pi}{2} , find \tan (A+B)

Answer:

Given \tan A = \frac{3}{4} and \cos B = \frac{9}{41} , where \pi < A  < \frac{3\pi}{2} and  0 < B < \frac{\pi}{2}

This means that A is in Quadrant III and B is in Quadrant I

Therefore, \sin B = -\sqrt{1 - \cos^2 B} =  \sqrt{1 - \Big( \frac{9}{41} \Big)^2 } =  \sqrt{\frac{1600}{1681}} = \frac{40}{41}

\therefore \tan B = \frac{\sin B}{\cos B} = \frac{\frac{40}{41}}{\frac{9}{41}} = \frac{40}{9}

\therefore \tan (A+B) = \frac{\tan A + \tan B}{1- \tan A \tan B} = \frac{\frac{3}{4} + \frac{40}{9}}{1- \frac{3}{4} \times \frac{40}{9}} = \frac{27+ 160}{36-120} = - \frac{187}{84}

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Question 5: If \sin A = \frac{1}{2} and \cos B = \frac{12}{13} , where \frac{\pi}{2} < A  < \pi   and  \frac{3\pi}{2} < B < 2\pi , find \tan (A-B)

Answer:

Given \sin A = \frac{1}{2} and \cos B = \frac{12}{13} , where \frac{\pi}{2} < A  < \pi   and  \frac{3\pi}{2} < B < 2\pi

This means that A is in Quadrant II and B is in Quadrant IV

Therefore, \cos A = -\sqrt{1 - \sin^2 A} = - \sqrt{1 - \Big( \frac{1}{2} \Big)^2 } = - \sqrt{\frac{3}{4}} = - \frac{\sqrt{3}}{2}

Similarly, \sin B = -\sqrt{1 - \cos^2 B} = - \sqrt{1 - \Big( \frac{12}{13} \Big)^2 } = - \sqrt{\frac{25}{169}} = - \frac{5}{13}

\therefore \tan A = \frac{\sin A}{\cos A} = \frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}} = - \frac{1}{\sqrt{3}}

Similarly, \tan B = \frac{\sin B}{\cos B} = \frac{\frac{-5}{13}}{\frac{12}{13}} = - \frac{5}{12}

\therefore \tan (A-B) = \frac{\tan A - \tan B}{1+ \tan A \tan B} = \frac{\frac{-1}{\sqrt{3}} - \frac{-5}{12}}{1+ \frac{-1}{\sqrt{3}} \times \frac{-5}{12}} = \frac{-12+ 5\sqrt{3}}{12\sqrt{3}+5} =  \frac{5\sqrt{3} - 12}{12\sqrt{3}+5}

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Question 6: If \sin A = \frac{1}{2} and \cos B = \frac{\sqrt{3}}{2} , where \frac{\pi}{2} < A  < \pi   and 0 < B < \frac{\pi}{2} , find the following:   i) \tan (A+B)    ii) \tan (A-B)

Answer:

Given \sin A = \frac{1}{2} and \cos B = \frac{\sqrt{3}}{2} , where \frac{\pi}{2} < A  < \pi   and 0 < B < \frac{\pi}{2}

This means that A is in Quadrant II and B is in Quadrant I

Therefore, \cos A = -\sqrt{1 - \sin^2 A} = - \sqrt{1 - \Big( \frac{1}{2} \Big)^2 } = - \sqrt{\frac{3}{4}} = - \frac{\sqrt{3}}{2}

Similarly, \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \Big( \frac{\sqrt{3}}{2} \Big)^2 } = \sqrt{\frac{1}{4}} = \frac{1}{2}

\therefore \tan A = \frac{\sin A}{\cos A} = \frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}} = - \frac{1}{\sqrt{3}}

Similarly, \tan B = \frac{\sin B}{\cos B} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}

i) \tan (A+B) = \frac{\tan A + \tan B}{1- \tan A \tan B} = \frac{\frac{-1}{\sqrt{3}}+ \frac{1}{\sqrt{3}}}{1 - \frac{-1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}} = \frac{-1+1}{1+\frac{1}{3}} = 0

ii) \tan (A-B) = \frac{\tan A - \tan B}{1+ \tan A \tan B} = \frac{\frac{-1}{\sqrt{3}}- \frac{1}{\sqrt{3}}}{1 + \frac{-1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}} = \frac{-2\sqrt{3}}{2} =  -\sqrt{3}

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Question 7: Evaluate the following:

i) \sin 78^o \cos 18^o - \cos 78^o \sin 18^o    ii) \cos 47^o \cos 13^o - \sin 47^o \sin 13^o

iii) \sin 36^o \cos 9^o + \cos 36^o \sin 9^o    iv)\cos 80^o \cos 20^o + \sin 80^o \sin 20^o

Answer:

i) \sin 78^o \cos 18^o - \cos 78^o \sin 18^o = \sin (78^o-18^o) = \sin 60^o = \frac{\sqrt{3}}{2}

ii) \cos 47^o \cos 13^o - \sin 47^o \sin 13^o = \cos (47^o+13^o) = \cos 60^o = \frac{1}{2}

iii) \sin 36^o \cos 9^o + \cos 36^o \sin 9^o = \sin ( 36^o + 9^o) = \sin 45^o = \frac{1}{\sqrt{2}}

iv)\cos 80^o \cos 20^o + \sin 80^o \sin 20^o = \cos (80^o-20^o) = \cos 60^o = \frac{1}{2}

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Question 8: If \cos A = - \frac{12}{13} and \cot B = \frac{24}{7} , where A lies in the second quadrant and B in the third quadrant, find the values of the following:

i) \sin (A+B)    ii) \cos (A+B)    iii) \tan (A + B)

Answer:

Given \cos A = - \frac{12}{13} and \cot B = \frac{24}{7} , where A lies in the second quadrant and B in the third quadrant

Therefore, \sin A = \sqrt{1 - \cos^2 A} =  \sqrt{1 - \Big( \frac{-12}{13} \Big)^2 } =  \sqrt{\frac{25}{169}} =  \frac{\sqrt{5}}{13}

\sin B = - \frac{1}{\sqrt{1+ \cot^2 B}}  = -  \frac{1}{\sqrt{1+ \Big( \frac{24}{7} \Big)^2}}  = -  \frac{1}{\sqrt{\frac{625}{49}}}  = -  \frac{7}{25}

i) \sin ( A+B) = \sin A \cos B + \cos A \sin B = \frac{5}{13} \times \frac{-24}{25} + \frac{-12}{13} \times \frac{-7}{25} = \frac{-120+84}{325} = \frac{-36}{325}

ii) \cos ( A+B) = \cos A \cos B - \sin A \sin B = \frac{-12}{13} \times \frac{-24}{25} - \frac{5}{13} \times \frac{-7}{25} = \frac{288+35}{325} = \frac{323}{325}

\tan ( A+B) = \frac{\sin (A+B)}{\cos (A+B) } = \frac{\frac{-36}{325}}{\frac{323}{325} }   = \frac{-36}{323}

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Question 9: Prove that \cos \frac{7\pi}{12} + \cos \frac{\pi}{12} = \sin \frac{5\pi}{12} - \sin \frac{\pi}{12}

Answer:

LHS = \cos \frac{7\pi}{12} + \cos \frac{\pi}{12}

= \cos \Big( \frac{\pi}{2} + \frac{\pi}{12} \Big) + \cos \Big( \frac{\pi}{2} - \frac{5\pi}{12} \Big)

= \sin \frac{5\pi}{12} - \sin \frac{\pi}{12}

= RHS. Hence proved.

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Question 10: Prove that \frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\sin (A+B)}{\sin (A-B)}

Answer:

LHS = \frac{\tan A + \tan B}{\tan A - \tan B}

= \frac{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}}{\frac{\sin A}{\cos A} - \frac{\sin B}{\cos B}}

= \frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B}

= \frac{\sin ( A+B)}{\sin ( A-B)}

= RHS. Hence proved.

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Question 11: 

i) \frac{\cos 11^o + \sin 11^o}{\cos 11^o - \sin 11^o} = \tan 56^o

ii) \frac{\cos 9^o + \sin 9^o}{\cos 9^o - \sin 9^o} = \tan 54^o

iii) \frac{\cos 8^o - \sin 8^o}{\cos 8^o + \sin 8^o} = \tan 37^o

Answer:

i) \frac{\cos 11^o + \sin 11^o}{\cos 11^o - \sin 11^o} = \frac{1+\tan 11^o}{1- \tan 11^o} = \frac{\tan 45^o + \tan 11^o}{1 - \tan 45^o \tan 11^o} = \tan (45^o + 11^o) = \tan 56^o

ii) \frac{\cos 9^o + \sin 9^o}{\cos 9^o - \sin 9^o} = \frac{1+\tan 9^o}{1- \tan 9^o} = \frac{\tan 45^o + \tan 9^o}{1 - \tan 45^o \tan 9^o} = \tan (45^o + 9^o) = \tan 54^o

iii) \frac{\cos 8^o - \sin 8^o}{\cos 8^o +\sin 8^o} = \frac{1-\tan 8^o}{1+ \tan 8^o} = \frac{\tan 45^o - \tan 8^o}{1 + \tan 45^o \tan 8^o} = \tan (45^o - 8^o) = \tan 37^o

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Question 12:

i) \sin \Big( \frac{\pi}{3} - x \Big) \cos \Big( \frac{\pi}{6} + x \Big) + \cos \Big( \frac{\pi}{3} - x \Big) \sin \Big( \frac{\pi}{6} + x \Big)  = 1

ii) \sin \Big( \frac{4\pi}{9} + 7 \Big) \cos \Big( \frac{\pi}{9} + 7 \Big) - \cos \Big( \frac{4\pi}{9} +7 \Big) \sin \Big( \frac{\pi}{9} + 7 \Big)  = \frac{\sqrt{3}}{2}

iii) \sin \Big( \frac{3\pi}{8} - 5 \Big) \cos \Big( \frac{\pi}{8} + 5 \Big) + \cos \Big( \frac{3\pi}{8} -5 \Big) \sin \Big( \frac{\pi}{9} + 5 \Big)  = 1

Answer:

i) LHS = \sin \Big( \frac{\pi}{3} - x \Big) \cos \Big( \frac{\pi}{6} + x \Big) + \cos \Big( \frac{\pi}{3} - x \Big) \sin \Big( \frac{\pi}{6} + x \Big)

Note: \sin (A + B) = \sin A \cos B + \cos A + \sin B

= \sin \Big[ \Big( \frac{\pi}{3} - x \Big) + \Big( \frac{\pi}{6} + x \Big) \Big]

= \sin \Big[ \frac{\pi}{3} + \frac{\pi}{6} \Big]

= \sin \frac{\pi}{2}

= 1 = RHS. Hence proved.

ii) LHS = \sin \Big( \frac{4\pi}{9} + 7 \Big) \cos \Big( \frac{\pi}{9} + 7 \Big) - \cos \Big( \frac{4\pi}{9} +7 \Big) \sin \Big( \frac{\pi}{9} + 7 \Big)

Note: \sin (A - B) = \sin A \cos B - \cos A + \sin B

= \sin \Big[ \Big( \frac{4\pi}{9} + 7 \Big) - \Big( \frac{\pi}{9} + 7 \Big) \Big]

= \sin \Big[ \frac{4\pi}{9} - \frac{\pi}{9} \Big]

= \sin \frac{3\pi}{9}

= \frac{\sqrt{3}}{2} RHS. Hence proved.

iii) LHS = \sin \Big( \frac{3\pi}{8} - 5 \Big) \cos \Big( \frac{\pi}{8} + 5 \Big) + \cos \Big( \frac{3\pi}{8} -5 \Big) \sin \Big( \frac{\pi}{9} + 5 \Big)

Note: \sin (A + B) = \sin A \cos B + \cos A + \sin B

= \sin \Big[ \Big( \frac{3\pi}{8} - 5 \Big) + \Big( \frac{\pi}{8} + 5 \Big) \Big]

= \sin \Big[ \frac{3\pi}{8} + \frac{\pi}{8} \Big]

= \sin \frac{\pi}{2}

= 1 = RHS. Hence proved.

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Question 13: Prove that: \frac{\tan 69^o + \tan 66^o}{1 - \tan 69^o \tan 66^o} =- 1

Answer:

LHS = \frac{\tan 69^o + \tan 66^o}{1 - \tan 69^o \tan 66^o}

Note: Since \tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

= \tan (69^o+66^o) = \tan 135^o - \cot 45^o = -1 = RHS. Hence proved.

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Question 14: i) If \tan A = \frac{5}{6} and \tan B = \frac{1}{11} , prove that A + B = \frac{\pi}{4}

ii) If \tan A = \frac{m}{m-1} and \tan B = \frac{1}{2m-1 } , prove that A - B = \frac{\pi}{4}

Answer:

i)  \tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

= \frac{\frac{5}{6} + \frac{1}{11}}{1 - \frac{5}{6} . \frac{1}{11}}   = \frac{55 + 6}{66-5}   = \frac{61}{61}

= 1 = \tan \frac{\pi}{4}

\therefore A+B = \frac{\pi}{4}

ii)  \tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

= \frac{\frac{m}{m-1} - \frac{1}{2m-1 }}{1 + \frac{m}{m-1} . \frac{1}{2m-1 }}

= \frac{m(2m-1) - (m-1)}{(m-1)(2m-1) + m}

= \frac{2m^2 - m - m + 1}{2m^2 - m - 2m + 1 + m}

= \frac{2m^2 - 2m + 1}{2m^2 - 2m + 1}

= 1 = \tan \frac{\pi}{4}

\therefore A-B = \frac{\pi}{4}

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Question 15: Prove that

i) \cos^2 \frac{\pi}{4} - \sin^2 \frac{\pi}{12} = \frac{\sqrt{3}}{4}

ii) \sin^2 (n+1) A - \sin^2 nA = \sin(2n+1) A \sin A

Answer:

i) LHS = \cos^2 \frac{\pi}{4} - \sin^2 \frac{\pi}{12}

= \Big( \frac{1}{\sqrt{2}} \Big)^2   - \sin^2 \frac{\pi}{12}

Lets calculate \sin^2 \frac{\pi}{12} = \sin^2 15^o

\sin 15^o = \sin ( 45^o- 30^o)

= \sin 45^o \cos 30^o - \cos 45^o \sin 30^o

= \Big( \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2} \Big)

= \frac{\sqrt{3} -1}{8}

Substituting it back

= \frac{1}{2} - \Big(  \frac{\sqrt{3} -1}{8}  \Big)

= \frac{1}{2} - \Big(  \frac{(3 +1 - 2 \sqrt{3})^2}{8} \Big)

= \frac{1}{2} - \Big( \frac{4 - 2 \sqrt{3}}{8} \Big)

= \frac{1}{2} - \Big( \frac{2 -  \sqrt{3}}{4} \Big)

= \frac{2 - 2 + \sqrt{3}}{4}

= \frac{\sqrt{3}}{4}

= RHS. Hence proved.

ii)  LHS = \sin^2 (n+1) A - \sin^2 nA

Since: \sin^2 A - \sin^2 B = \sin (A+B) \sin(A-B)

 = \sin [ (n+1)A + nA ] \sin [(n+1)A - nA ]

 = \sin (2nA + A) \sin A

= \sin (2n+1)A \sin A

 = RHS. Hence Proved.

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Question 16: Prove that:

i) \frac{\sin (A+B) + \sin (A-B)}{\cos (A+B) + \cos (A-B)} = \tan A

ii) \frac{\sin (A-B)}{\cos A \cos B} + \frac{\sin (B-C)}{\cos B \cos C} + \frac{\sin (C-A)}{\cos C \cos A}   = 0

iii) \frac{\sin (A-B)}{\sin A \sin B} + \frac{\sin (B-C)}{\sin B \sin C} + \frac{\sin (C-A)}{\sin C \sin A}   = 0

iv) \sin^2 B= \sin^2 A + \sin^2 (A-B) - 2 \sin A \cos B \sin (A-B)

v) \cos^2 A + \cos^2 B - 2 \cos A \cos B \cos (A+ B) = \sin^2 (A+B)

vi) \frac{\tan (A+B)}{\cot(A-B)} = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}

Answer:

i) LHS = \frac{\sin (A+B) + \sin (A-B)}{\cos (A+B) + \cos (A-B)}

= \frac{2 \sin A \cos B}{2 \cos A \cos B} = \tan A = RHS. Hence proved.

ii) LHS = \frac{\sin (A-B)}{\cos A \cos B} + \frac{\sin (B-C)}{\cos B \cos C} + \frac{\sin (C-A)}{\cos C \cos A}

= \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B} + \frac{\sin B \cos C - \cos B \sin C}{\cos B \cos C} + \frac{\sin C \cos A - \cos C \sin A}{\cos C \cos A}

= \tan A - \tan B + \tan B - \tan C + \tan C - \tan A

= 0 = RHS. Hence proved.

iii) LHS = \frac{\sin (A-B)}{\sin A \sin B} + \frac{\sin (B-C)}{\sin B \sin C} + \frac{\sin (C-A)}{\sin C \sin A}

= \frac{\sin A \cos B - \cos A \sin B}{\sin A \sin B} + \frac{\sin B \cos C - \cos B \sin C}{\sin B \sin C} + \frac{\sin C \cos A - \cos C \sin A}{\sin C \sin A}

= \cot B - \cot A + \cot C - \cot B + \cot A - \cot C

= 0 = RHS. Hence proved.

iv) RHS = \sin^2 A + \sin^2 (A-B) - 2 \sin A \cos B \sin (A-B)

= \sin^2 A + \sin (A_B) [ \sin (A-B) - 2 \sin A \cos B ]

= \sin^2 A + \sin (A - B) [ \sin A \cos B - \cos A \sin B - 2 \sin A \cos B ]

= \sin^2 A - \sin (A - B)  [ \sin A \cos B + \cos A \sin B ]

= \sin^2 A - \sin (A - B) \sin ( A + B)

= \sin^2 - ( \sin^2 A - \sin^2 B)

= \sin^2 B = LHS. Hence Proved

v) LHS = \cos^2 A + \cos^2 B - 2 \cos A \cos B \cos (A+ B)

= \cos^2 + [1 - \sin^2 B ] - 2 \cos A \cos B \cos (A+B)

= (\cos^2 A - \sin^2 B) - 2 \cos A \cos B \cos (A+B) + 1

= \cos (A+B) \cos (A - B) - 2 \cos A \cos B \cos ( A + B) + 1

= \cos ( A+B) [ \cos (A - B) - 2 \cos A \cos B ] + 1

= \cos ( A + B) [ \cos A \cos B + \sin A \sin B - 2 \cos A \cos B ] + 1

= \cos (A+B) [ - \cos A \cos B + \sin A \sin B ] + 1

= - \cos (A + B) [\cos A \cos B - \sin A \sin B ] + 1

= - \cos (A + B) \cos ( A + B) + 1

= - \cos^2 ( A + B) + 1 = \sin^2 ( A + B) = RHS. Hence Proved.

vi) LHS = \frac{\tan (A+B)}{\cot(A-B)}

=  \frac{\tan ( A + B)}{\frac{1}{\tan ( A - B)}}

= \tan ( A + B ) \tan ( A - B)

= \Big[ \frac{\tan A + \tan B}{1 - \tan A \tan B} \Big] + \Big[ \frac{\tan A - \tan B}{1 + \tan A \tan B} \Big]

= \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B} = RHS. Hence proved.

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Question 17: Prove that:

i) \tan 8x - \tan 6x - \tan 2x = \tan 8x \tan 6x \tan 2x

ii) \tan \frac{\pi}{12} + \tan \frac{\pi}{6} + \tan \frac{\pi}{12} \tan \frac{\pi}{6} = 1

iii) \tan 36^o + \tan 9^o + \tan 36^o \tan 9^o = 1

iv) \tan 13x - \tan 9x  - \tan 4x = \tan 13x \tan 9x \tan 4x

Answer:

i) 8x = 6x + 2x

\Rightarrow  \tan 8x = \tan (6x + 2x)

\Rightarrow \tan 8x = \frac{\tan 6x + \tan 2x}{1- \tan 6x \tan 2x}

\Rightarrow \tan 8x ( 1- \tan 6x \tan 2x ) = \tan 6x + \tan 2x

\Rightarrow \tan 8x - \tan 8x \tan 6x \tan 2x = \tan 6x + \tan 2x

\Rightarrow \tan 8x - \tan 6x - \tan 2x = \tan 8x \tan 6x \tan 2x

ii) \frac{\pi}{4} = \frac{\pi}{6} + \frac{\pi}{12}

\Rightarrow  \tan \frac{\pi}{4} = \tan (\frac{\pi}{6} + \frac{\pi}{12})

\Rightarrow \tan \frac{\pi}{4} = \frac{\tan \frac{\pi}{6}+ \tan \frac{\pi}{12}}{1- \tan \frac{\pi}{6} \tan \frac{\pi}{12}}

\Rightarrow 1 ( 1- \tan \frac{\pi}{6} \tan \frac{\pi}{12} ) = \tan \frac{\pi}{6} + \tan \frac{\pi}{12}

\Rightarrow 1 -  \tan \frac{\pi}{6} \tan \frac{\pi}{12} = \tan \frac{\pi}{6} + \tan \frac{\pi}{12}

\Rightarrow 1 =  \tan \frac{\pi}{6} + \tan \frac{\pi}{12} +  \tan \frac{\pi}{6} \tan \frac{\pi}{12}

iii)  45^o = 36^o + 9^o

\Rightarrow  \tan 45^o = \tan (36^o + 9^o)

\Rightarrow \tan 45^o = \frac{\tan 36^o + \tan 9^o}{1- \tan 36^o \tan 9^o}

\Rightarrow 1 ( 1- \tan 36^o \tan 9^o ) = \tan 36^o + \tan 9^o

\Rightarrow 1 -  \tan 36^o \tan 9^o = \tan 36^o + \tan 9^o

\Rightarrow 1 - \tan 36^o - \tan 9^o =  \tan 36^o \tan 9^o

iv)  13x = 9x + 4x

\Rightarrow  \tan 13x = \tan (9x + 4x)

\Rightarrow \tan 13x = \frac{\tan 9x + \tan 4x}{1- \tan 9x \tan 4x}

\Rightarrow \tan 13x ( 1- \tan 9x \tan 4x ) = \tan 9x + \tan 4x

\Rightarrow \tan 13x - \tan 13x \tan 9x \tan 4x = \tan 9x + \tan 4x

\Rightarrow \tan 13x - \tan 9x - \tan 4x = \tan 13x \tan 9x \tan 4x

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Question 18: Prove that \frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} = \tan 3x \tan x

Answer:

RHS = \tan 3x \tan x

= \tan (2x+x) \tan (2x-x)

= [ \frac{\tan 2x + \tan x}{1- \tan 2x \tan x} ] [  \frac{\tan 2x - \tan x}{1+ \tan 2x \tan x} ]

= \frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x}

= LHS. Hence proved.

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Question 19: \frac{\sin(x+y)}{\sin(x-y)} = \frac{a+b}{a-b} . show that \frac{\tan x}{\tan y} = \frac{a}{b}

Answer:

\frac{\sin(x+y)}{\sin(x-y)} = \frac{a+b}{a-b}

\Rightarrow \frac{\sin x \cos y + \cos x \sin y}{\sin x \cos y - \cos x \sin y} = \frac{a+b}{a-b}

Using Componendo and Dividendo

\Rightarrow \frac{\sin x \cos y + \cos x \sin y + \sin x \cos y - \cos x \sin y}{\sin x \cos y + \cos x \sin y - \sin x \cos y + \cos x \sin y} = \frac{a+b + a - b}{a + b - a+b}

\Rightarrow \frac{2\sin x \cos y}{2\cos x \sin y} = \frac{2a}{2b}

\Rightarrow \frac{\sin x \cos y}{\cos x \sin y} = \frac{a}{b}

\Rightarrow \frac{\tan x}{\tan y} = \frac{a}{b}

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Question 20: If \tan A = x \tan B , prove that \frac{\sin(A-B)}{\sin(A+B)} = \frac{x-1}{x+1}

Answer:

\tan A = x \tan B

\frac{\sin A}{\cos A} = x \frac{\sin A}{\cos A}

\sin A \cos B = x \sin B \cos A

Now \frac{\sin (A-B)}{\sin (A+B)} = \frac{\sin A \cos B - \cos A \sin B}{\sin A \cos B + \cos A \sin B}

\frac{\sin (A-B)}{\sin (A+B)} = \frac{x \cos A \sin B - \cos A \sin B}{x \cos A \sin B - \cos A \sin B}

\frac{\sin (A-B)}{\sin (A+B)} = \frac{(x -1) \cos A \sin B}{(x - 1 ) \cos A \sin B }

\frac{\sin (A-B)}{\sin (A+B)} = \frac{(x -1) }{(x - 1 )  }

Hence proved.

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Question 21: If \tan (A+B) = x and \tan (A-B) = y , find the value of \tan 2A and \tan 2B

Answer:

Given \tan (A+B) = x and \tan (A-B) = y

\tan 2A = \tan [ (A+B) + (A -B) ]

= \frac{\tan (A+B)+ \tan (A-B)}{1 - \tan (A+B) \tan (A-B)}

= \frac{x+y}{1-xy}

Similarly,

\tan 2B = \tan [ (A+B) - (A -B) ]

= \frac{\tan (A+B) - \tan (A-B)}{1 + \tan (A+B) \tan (A-B)}

= \frac{x-y}{1+xy}

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Question 22: If \cos A + \sin B = m and \sin A + \cos B = n , prove that 2 \sin (A+B) = m^2 + n^2 -2

Answer:

Given \cos A + \sin B = m and \sin A + \cos B = n

m^2 + n^2 - 2

= (\cos A + \sin B)^2 + (\sin A + \cos B)^2 - 2

= \cos^2 A + \sin^2 B + 2 \cos A \sin B + \sin^2 A + \cos^2 B + 2 \sin A \cos B - 2

= ( \cos^2 A + \sin^2 A) + ( \cos^2 B + \sin^2 B) + 2 (\cos A \sin B + \sin A \cos B) - 2

= 2 + + 2 (\cos A \sin B + \sin A \cos B) - 2

= 2 (\cos A \sin B + \sin A \cos B)

= 2 \sin ( A + B)    Hence proved.

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Question 23: If \tan A + \tan B = a and \cot A + \cot B = b , prove that: \cot (A+B) = \frac{1}{a} - \frac{1}{b}

Answer:

Given \tan A + \tan B = a and \cot A + \cot B = b

\cot A + \cot B = b

\Rightarrow \frac{1}{\tan A} + \frac{1}{\tan B} = b

\Rightarrow \frac{\tan A + \tan B}{1 - \tan A \tan B} = b

\Rightarrow \tan A \tan B = \frac{a}{b}

\because \cot ( A + B) = \frac{1}{\tan ( A + B)}

= \frac{1 - \tan A \tan B}{\tan A + \tan B}

= \frac{1 - \frac{a}{b}}{a}

= \frac{b-a}{ab}

= \frac{1}{a} - \frac{1}{b}

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Question 24: If x lies in the first quadrant and \cos x = \frac{8}{17} , then prove that

\cos \Big( \frac{\pi}{6} + x \Big) + \cos \Big( \frac{\pi}{4} - x \Big) + \cos \Big( \frac{2\pi}{3} - x \Big)  = \Big( \frac{\sqrt{3}-1}{2} + \frac{1}{\sqrt{2}} \Big) \frac{23}{17}

Answer:

Given \cos x = \frac{8}{17}

\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \Big( \frac{8}{17} \Big)^2 } = \sqrt{\frac{225}{289}} = \frac{15}{17}

\cos \Big( \frac{\pi}{6} + x \Big) + \cos \Big( \frac{\pi}{4} - x \Big) + \cos \Big( \frac{2\pi}{3} - x \Big)

= \Big[ \cos \frac{\pi}{6} \cos x - \sin \frac{\pi}{6} \sin x \Big]  + \Big[ \cos \frac{\pi}{4} \cos x + \sin \frac{\pi}{4} \sin x \Big] + \Big[ \cos \frac{2\pi}{3} \cos x + \sin \frac{2\pi}{3} \sin x \Big]

= \Big[ \cos \frac{\pi}{6} + \cos \frac{\pi}{4} + \cos \frac{2\pi}{3} \Big] \cos x  + \Big[ - \sin \frac{\pi}{6} + \sin \frac{\pi}{4} + \sin \frac{2\pi}{3} \Big] \sin x

= \Big[ \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} + \cos \Big( \frac{\pi}{2}   + \frac{\pi}{6} \Big) \Big] \times \frac{8}{17} +  \Big[ - \frac{1}{2} + \frac{1}{\sqrt{2}} + \sin \Big( \frac{\pi}{2}   + \frac{\pi}{6} \Big) \Big] \times \frac{15}{17}

= \Big[ \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \sin \frac{\pi}{6} \Big] \times \frac{8}{17} +  \Big[ - \frac{1}{2} + \frac{1}{\sqrt{2}} + \cos \frac{\pi}{6} \Big] \times \frac{15}{17}

= \Big[ \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} - \frac{1}{2} \Big] \times \frac{8}{17} +  \Big[ - \frac{1}{2} + \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \Big] \times \frac{15}{17}

= \Big( \frac{\sqrt{3}-1}{2} + \frac{1}{\sqrt{2}}   \Big)  ( \frac{8}{17} + \frac{15}{17} )

= \Big( \frac{\sqrt{3}-1}{2} + \frac{1}{\sqrt{2}}   \Big)  \Big( \frac{23}{17} \Big)

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Question 25: If \tan x + \tan \Big( x + \frac{\pi}{3} \Big) +  \tan \Big( x + \frac{2\pi}{3} \Big) = 3 , then  prove that \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} = 1

Answer:

Given

\tan x + \tan \Big( x + \frac{\pi}{3} \Big) +  \tan \Big( x + \frac{2\pi}{3} \Big) = 3

\Rightarrow \tan x + \frac{\tan x + \tan \frac{\pi}{3} }{1 - \tan x \tan \frac{\pi}{3} } + \frac{\tan x + \tan \frac{2\pi}{3} }{1 - \tan x \tan \frac{2\pi}{3} } = 3

\Rightarrow \tan x + \frac{\tan x + \sqrt{3} }{1 - \sqrt{3}\tan x  } + \frac{\tan x + \tan \Big( \frac{\pi}{2} + \frac{\pi}{3}  \Big) }{1 - \tan x \tan \Big( \frac{\pi}{2} + \frac{\pi}{3}  \Big) } = 3

\Rightarrow \tan x + \frac{\tan x + \sqrt{3} }{1 - \sqrt{3}\tan x  } + \frac{\tan x - \cot  \frac{\pi}{3}  }{1 - \tan x \cot  \frac{\pi}{3} } = 3

\Rightarrow \tan x + \frac{\tan x + \sqrt{3} }{1 - \sqrt{3}\tan x  } + \frac{\tan x - \sqrt{3}  }{1 + \sqrt{3}\tan x  } = 3

\Rightarrow \tan x + \frac{(\tan x + \sqrt{3})(1 + \sqrt{3}\tan x) + (\tan x - \sqrt{3})(1 - \sqrt{3}\tan x)}{(1 - \sqrt{3}\tan x )(1 + \sqrt{3}\tan x)} = 3

\Rightarrow \tan x + \frac{\tan x +\sqrt{3} + \sqrt{3} \tan^2 x + 3 \tan x + \tan x - \sqrt{3} - \sqrt{3} \tan^2 x + 3 \tan x }{1 - 3 \tan^2 x} = 3

\Rightarrow \tan x + \frac{8 \tan x}{1-3 \tan^2 x} = 3 

\Rightarrow \frac{\tan x (1-3 \tan^2 x) + 8 \tan x }{1-3 \tan^2 x} = 3

\Rightarrow \frac{\tan x - 3 \tan^2 x + 8 \tan x }{1-3 \tan^2 x} = 3

\Rightarrow \frac{9 \tan x - 3 \tan^3 x }{1-3 \tan^2 x} = 3

\Rightarrow \frac{3 \tan x - \tan^3 x }{1-3 \tan^2 x} = 1 . Hence proved.

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Question 26: If  \sin (\alpha + \beta) = 1 and \sin (\alpha - \beta) = \frac{1}{2} , where 0 \leq \alpha, \beta \leq \frac{\pi}{2} , then find the values of \tan ( \alpha + 2 \beta) and \tan ( 2\alpha +  \beta)

Answer:

Given:

\sin (\alpha + \beta) = 1  \Rightarrow \alpha + \beta = \frac{\pi}{2}    … … … i)

\sin (\alpha - \beta) = \frac{1}{2}  \Rightarrow \alpha - \beta = \frac{\pi}{6}    … … … ii)

Solving i) and ii) we get

2 \alpha = \frac{\pi}{2} + \frac{\pi}{6} \Rightarrow \alpha = \frac{\pi}{3}

Also \beta = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}

\tan ( \alpha + 2 \beta) = \tan ( \frac{\pi}{3} + 2 \times \frac{\pi}{6} )  = \tan \frac{\pi}{3} =  \tan ( \frac{\pi}{2} + \times \frac{\pi}{6} ) = - \cot \frac{\pi}{6} = -\sqrt{3} 

\tan ( 2\alpha +  \beta) = \tan (2 \frac{\pi}{3} + \frac{\pi}{6} )  = \tan \frac{5\pi}{6} =  \tan ( \frac{\pi}{2} + \times \frac{\pi}{3} ) = - \cot \frac{\pi}{3} = \frac{-1}{\sqrt{3}} 

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Question 27: If \alpha and \beta are two different values of x lying between 0 and 2\pi which satisfy the equation 6 \cos x + 8 \sin x = 9 , find the value of \sin (\alpha + \beta)

Answer:

Given

6 \cos x + 8 \sin x = 9

Eliminating \sin x

\Rightarrow 8 \sin x = 9 - 6 \cos x

Squaring both sides

\Rightarrow (8 \sin x)^2 = (9 - 6 \cos x)^2

\Rightarrow 64 \sin^2 x = 81 + 36 \cos^2 x - 108 \cos x

\Rightarrow 64 (1 - \cos^2 x) = 81 + 36 \cos^2 x - 108 \cos x

\Rightarrow 64 - 64 \cos^2 x = 81 + 36 \cos^2 x - 108 \cos x

\Rightarrow 100 \cos^2 x - 108 \cos x + 17 = 0

If \alpha and \beta are roots of equation

\Rightarrow \cos \alpha \times \cos \beta = \frac{17}{100}

Similarly

6 \cos x + 8 \sin x = 9

Eliminating \cos x

\Rightarrow 6 \cos x = 9 - 8 \sin x

Squaring both sides

\Rightarrow  (6 \cos x)^2 = (9 - 8 \sin x)^2

\Rightarrow 36 \cos^2 x = 81 + 64 \sin^2 x - 144 \sin x

\Rightarrow 36 (1 - \sin^2 x) = 81 + 64 \sin^2 x - 144 \sin x

\Rightarrow 36 - 36 \sin^2 x = 81 + 64 \sin^2 x - 144 \sin x

\Rightarrow 100 \sin^2 x - 144 \sin x + 17 = 0

If \alpha and \beta are roots of equation

\Rightarrow \sin \alpha \times \sin \beta = \frac{17}{100}

Now \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \frac{17}{100} - \frac{45}{100} = \frac{-28}{100} = \frac{-7}{25}

\sin (\alpha + \beta) = \sqrt{1 - [\cos(\alpha + \beta) ]^2} = \sqrt{1 - \Big[ \frac{-7}{25} \Big]^2} = \sqrt{1 - \frac{49}{625} } = \sqrt{\frac{576}{625}} = \frac{24}{25}

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Question 28: If \sin \alpha + \sin \beta = a and \cos \alpha + \cos \beta = b , show that

i) \sin (\alpha + \beta) = \frac{2ab}{a^2 + b^2}

ii) \cos (\alpha + \beta) = \frac{b^2 - a^2}{b^2 + a^2}

Answer:

Given, \sin \alpha + \sin \beta = a and \cos \alpha + \cos \beta = b

\therefore a^2 + b^2 = (\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2

\Rightarrow a^2 + b^2= \sin^2 \alpha +  \sin^2 \beta  + 2 \sin \alpha \sin \beta +  \cos^2 \alpha +  \cos^2 \beta  + 2 \cos \alpha \cos \beta

\Rightarrow a^2 + b^2= 2 + 2 (\sin \alpha \sin \beta + \cos \alpha \cos \beta)

\Rightarrow a^2 + b^2= 2 + 2 \cos ( \alpha + \beta )

Similarly, \therefore b^2 - a^2 = (\cos \alpha + \cos \beta)^2 - (\sin \alpha + \sin \beta)^2   

\Rightarrow b^2 - a^2 =  \cos^2 \alpha +  \cos^2 \beta  + 2 \cos \alpha \cos \beta - \sin^2 \alpha -  \sin^2 \beta  - 2 \sin \alpha \sin \beta 

\Rightarrow b^2 - a^2 = (\cos^2 \alpha -  \sin^2 \beta) + (\cos^2 \beta - \sin^2 \alpha) + 2( cos \alpha \cos \beta -  \sin \alpha \sin \beta)

\Rightarrow b^2 - a^2 = \cos (\alpha + \beta) \cos(\alpha - \beta) + \cos ( \beta + \alpha) \cos ( \beta - \alpha) + 2 \cos (\alpha + \beta)

\because \cos (\beta - \alpha) = \cos [ - ( \alpha - \beta ) ] =  \cos ( \alpha - \beta )

\Rightarrow b^2 - a^2 = \cos (\alpha + \beta) [ 2 \cos ( \alpha - \beta )  + 2]

\Rightarrow b^2 - a^2 = \cos (\alpha + \beta) [ a^2 + b^2 ]

\Rightarrow \cos ( \alpha + \beta )  = \frac{b^2 - a^2}{b^2 + a^2}

\sin ( \alpha + \beta ) = \sqrt{1 - [\cos ( \alpha + \beta )]^2 }

= \sqrt{1 - \Big( \frac{b^2 - a^2}{b^2 + a^2} \Big)^2 }

= \sqrt{ \frac{b^4 + a^4 + 2 b^2 a^2 - b^4 - a^4 + 2 b^2a^2}{(b^2 + a^2)^2} }

= \frac{2ba}{b^2 + a^2}

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Question 29: Prove that:

i) \frac{1}{\sin (x-a) \sin (x-b) } = \frac{\cot (x-a) - \cot (x-b)}{\sin (a-b)}

ii) \frac{1}{\sin (x-a) \cos (x-b) } = \frac{\cot (x-a) + \tan (x-b)}{\cos (a-b)}

iii) \frac{1}{\cos (x-a) \cos (x-b) } = \frac{\tan (x-b) - \tan (x-a)}{\sin (a-b)}

Answer:

i) LHS = \frac{1}{\sin (x-a) \sin (x-b) }

= \frac{1}{\sin(a-b)} \Big[  \frac{\sin(a-b)}{\sin (x-a) \sin (x-b)}   \Big]

= \frac{1}{\sin(a-b)} \Big[  \frac{\sin [(x-b) - (x-a) ] }{\sin (x-a) \sin (x-b)}   \Big]

= \frac{1}{\sin(a-b)} \Big[  \frac{\sin (x-b) \cos (x-a) - \cos (x-b) \sin (x-a)  }{\sin (x-a) \sin (x-b)}   \Big]

= \frac{1}{\sin(a-b)} \Big[  \frac{\sin (x-b) \cos (x-a)}{\sin (x-a) \sin (x-b)} - \frac{\cos (x-b) \sin (x-a)}{\sin (x-a) \sin (x-b)}   \Big]

= \frac{1}{\sin(a-b)} [ \cot (x-a) - \cot ( x- b) ]

= \frac{ \cot (x-a) - \cot ( x- b)}{\sin(a-b)}

= RHS. Hence proved.

ii) LHS = \frac{1}{\sin (x-a) \cos (x-b) }

= \frac{1}{\cos(a-b)} \Big[  \frac{\cos(a-b)}{\sin (x-a) \cos (x-b)}   \Big]

= \frac{1}{\cos(a-b)} \Big[  \frac{\cos [(x-b) - (x-a) ] }{\sin (x-a) \cos (x-b)}   \Big]

= \frac{1}{\cos(a-b)} \Big[  \frac{\cos (x-b) \cos (x-a) + \sin (x-b) \sin (x-a)  }{\sin (x-a) \cos (x-b)}   \Big]

= \frac{1}{\cos(a-b)} \Big[  \frac{\cos (x-b) \cos (x-a)}{\sin (x-a) \cos (x-b)} + \frac{\sin (x-b) \sin (x-a)}{\sin (x-a) \cos (x-b)}   \Big]

= \frac{1}{\cos(a-b)} [ \cot (x-a) + \tan ( x- b) ]

= \frac{ \cot (x-a) + \tan ( x- b)}{\cos(a-b)}

= RHS. Hence proved.

iii) LHS = \frac{1}{\cos (x-a) \cos (x-b) }

= \frac{1}{\sin(a-b)} \Big[  \frac{\sin(a-b)}{\cos (x-a) \cos (x-b)}   \Big]

= \frac{1}{\sin(a-b)} \Big[  \frac{\sin [(x-b) - (x-a) ] }{\cos (x-a) \cos (x-b)}   \Big]

= \frac{1}{\sin(a-b)} \Big[  \frac{\sin (x-b) \cos (x-a) - \cos (x-b) \sin (x-a)  }{\cos (x-a) \cos (x-b)}   \Big]

= \frac{1}{\sin(a-b)} \Big[  \frac{\sin (x-b) \cos (x-a)}{\cos (x-a) \cos (x-b)} - \frac{\cos (x-b) \sin (x-a)}{\cos (x-a) \cos (x-b)}   \Big]

= \frac{1}{\sin(a-b)} [ \tan (x-b) - \tan ( x- a) ]

= \frac{ \tan (x-b) - \tan ( x- a)}{\sin(a-b)}

= RHS. Hence proved.

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Question 30: If \sin \alpha \sin \beta - \cos \alpha \cos \beta + 1 = 0 , prove that 1 + \cot \alpha \tan \beta = 0

Answer:

Given \sin \alpha \sin \beta - \cos \alpha \cos \beta + 1 = 0

\Rightarrow - ( \cos \alpha \cos \beta - \sin \alpha \sin \beta) = - 1

\Rightarrow  \cos ( \alpha + \beta)  =  1

\therefore \sin ( \alpha + \beta ) =  \sqrt{ 1 - \cos^2 ( \alpha + \beta ) } = \sqrt{ 1 - 1 } = 0

1 + \cot \alpha \tan \beta =  1 + \frac{\cos \alpha}{\sin \alpha} \times \frac{\sin \beta}{\cos \beta}

\Rightarrow  1 + \cot \alpha \tan \beta = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\sin \alpha \cos \beta  }

\Rightarrow  1 + \cot \alpha \tan \beta = \frac{\sin ( \alpha + \beta )}{\sin \alpha \cos \beta}

\Rightarrow  1 + \cot \alpha \tan \beta = 0      Hence Proved.

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Question 31: If \tan \alpha = x + 1 and \tan \beta = x - 1 , show that 2 \cot (\alpha - \beta) = x^2

Answer:

Given \tan \alpha = x + 1 and \tan \beta = x - 1

2 \cot (\alpha - \beta) = \frac{2}{\tan (\alpha - \beta)}

\Rightarrow 2 \cot (\alpha - \beta) = \frac{2 ( 1 + \tan \alpha \tan \beta}{\tan \alpha - \tan \beta}

\Rightarrow 2 \cot (\alpha - \beta) = 2 \Big[  \frac{1 + ( x + 1 ) ( x - 1 ) }{(x+ 1 ) - ( x - 1)} \Big]

\Rightarrow 2 \cot (\alpha - \beta) = 2 \Big[ \frac{1 +x^2 - 1}{2} \Big]

\Rightarrow 2 \cot (\alpha - \beta) = x^2 . Hence proved.

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Question 32: If  angle \theta is divided into two parts such that the tangents of one parts is \lambda times the tangent of other, and \phi is their difference, then show that \sin \theta = \frac{\lambda - 1}{\lambda + 1} \sin \phi

Answer:

Let \alpha and \beta be the two parts

\theta = \alpha + \beta

\phi = \alpha - \beta

\tan \alpha = \lambda \tan \beta

\Rightarrow \frac{\tan \alpha}{\tan \beta} = \frac{\lambda}{1}

Applying componendo and dividendo

\Rightarrow \frac{\tan \alpha + \tan \beta}{\tan \alpha - \tan \beta} = \frac{\lambda+ 1}{\lambda - 1}

\Rightarrow \frac{ \frac{\sin \alpha}{\cos \alpha} + \frac{\sin \beta}{\cos \beta}}{\frac{\sin \alpha}{\cos \alpha} - \frac{\sin \beta}{\cos \beta}} = \frac{\lambda+ 1}{\lambda - 1}

\Rightarrow \frac{\sin \alpha \cos \beta + \sin \beta \cos \alpha}{\sin \alpha \cos \beta - \sin \beta \cos \alpha} = \frac{\lambda+ 1}{\lambda - 1}

\Rightarrow \frac{\sin ( \alpha + \beta) }{\sin ( \alpha - \beta) } = \frac{\lambda+ 1}{\lambda - 1}

\Rightarrow \frac{\sin \theta }{\sin \phi} = \frac{\lambda+ 1}{\lambda - 1}

\Rightarrow \sin \theta = \Big( \frac{\lambda+ 1}{\lambda - 1} \Big) \sin \phi       Hence proved.

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Question 33: If \tan x = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} , then show that \sin \alpha + \cos \alpha = \sqrt{2} \cos x

Answer:

Given \tan x = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}

Dividing both numerator and denominator by \cos \alpha

\Rightarrow \tan \theta = \frac{\tan \alpha - 1}{\tan \alpha + 1}

\Rightarrow \tan \theta = \frac{\tan \alpha - \tan \frac{\pi}{4}}{1 + \tan \alpha \tan \frac{\pi}{4}}

\Rightarrow \tan \theta = \tan (\alpha  - \frac{\pi}{4} )

\Rightarrow \theta = \alpha  - \frac{\pi}{4}

\Rightarrow \cos \theta = \cos (\alpha  - \frac{\pi}{4} )

\Rightarrow \cos \theta = \cos \alpha \cos \frac{\pi}{4} + \sin \alpha \sin \frac{\pi}{4}

\Rightarrow \cos \theta = \frac{\cos \alpha}{\sqrt{2}} + \frac{\sin \alpha}{\sqrt{2}}

\Rightarrow  \sqrt{2}  \cos \theta = \cos \alpha + \sin \alpha      Hence proved.

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Question 34: If