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Introduction To Trigonometry Exercise 8.3


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Q.1     Evaluate :
           (i) {{\sin 18^\circ } \over {\cos 72^\circ }}
           (ii) {{\tan 26^\circ } \over {\cot 64^\circ }}
           (iii) cos 48° – sin 42°
           (iv) cosec 31° – sec 59°
Sol.       (i) {{\sin 18^\circ } \over {\cos 72^\circ }} = {{\sin \left( {90^\circ - 72} \right)} \over {\cos 72^\circ }} = {{\cos 72^\circ } \over {\cos 72^\circ }} = 1[Since,\sin 90\,\,-\theta = \cos \theta ]

              (ii) {{\tan 26^\circ } \over {\cot 64^\circ }} = {{\tan \left( {90^\circ - 64^\circ } \right)} \over {\cot 64^\circ }} = {{\cot 64^\circ } \over {\cot 64^\circ }} = 1 [Since,\tan \left( {90 - \theta } \right) = \cot \theta ]

              (iii) cos48° – sin42° = cos (90° – 42°) – sin 42°
                                             = sin 42° – sin 42° = 0 [Since,\cos \left( {90 - \theta } \right) = \sin \theta ]
              (iv) cosec 31° – sec 59° = cosec (90° – 59°) – sec 59° = 0
                                                   = \sec 59^\circ - \sec 59^\circ = 0 [Since,\cos ec\left( {60 - \theta } \right) = \sec \theta ]


Q.2     Show that
           (i) tan 48° tan 23° tan 42° tan 67° = 1
           (ii) cos 38° cos 52° – sin 38° sin 52 ° = 0
Sol.       (i) tan 48° tan 23° tan 42 ° tan 67°
              = tan (90° – 42°) tan(90° – 67°) tan 42° tan 67°
              = cot 42° cot 67° tan 42° tan 67°  [Since,\tan \left( {90 - \theta } \right) = \cot \theta ]
               = {1 \over {\tan 42^\circ }}.{1 \over {\tan 67^\circ }}.\tan 42^\circ \tan 67^\circ = 1

              (ii) cos 38°cos 52°– sin 38°sin 52°
              = cos(90° – 52°)cos(90°– 38°) – sin 38°sin 52°
              = sin 52° sin 38° – sin 38°sin 52° = 0 [Since,\cos \left( {90^\circ - \theta } \right) = \sin \theta ]


Q.3     If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Sol.       We are given that
             tan 2A = cot (A – 18°)...(1)
             Since tan 2A = cot (90° – 2A), so we can write (1) as
             cot(90°–2A) = cot (90° – 2A), so we can write (1) as
             cot (90° – 2A) = cot (A – 18°) [Since,\cot \left( {90^\circ - \theta } \right) = \tan \theta ]
             Since (90° – 2A) and (A – 18°) are both acute angle therefore,
             90° – 2A = A – 18°
              \Rightarrow – 2A – A = – 18° – 90°
              \Rightarrow – 3A = – 108°
              \Rightarrow A = 36°


Q.4     If tan A = cot B, prove that A + B = 90°.
Sol.       We are given that
              tan A = cot B...(1)
              Since tan A = cot(90° – A), so we can write (1) as
              cot (90° – A ) = cot B [Since,\cot \left( {90^\circ - \theta } \right) = \tan \theta ]
              Since (90° – A) and B are both acute angles, therefore,
              (90° – A) = B
               \Rightarrow A + B = 90°


Q.5     If sec 4 A = cosec (A – 20°), where 4 A is an acute angle, find the value of A.
Sol.      We are given that sec 4A = cosec (A – 20°). ....(1)
             Since sec 4A = cosec (90°– 4A), so we can write (1) as
             cosec (90° – 4A) = cosec (A – 20°) [Since,\cos ec\left( {90^\circ - \theta } \right) = \sec \theta ]
             Since (90° – 4A) and (A – 20°) are both acute angles, therefore,
             90° – 4A = A – 20°
              \Rightarrow – 4A – A = – 20° – 90°
              \Rightarrow – 5A = 110°
              \Rightarrow A = 22°


Q.6     If A, B and C are interior angles of a \Delta ABC, then show that
           \sin \left( {{{B + C} \over 2}} \right) = \cos {A \over 2}
Sol.       Since A, B and C are the interior angles of a \Delta ABC, therefore,
              A + B + C = 180°
               \Rightarrow {{A + B + C} \over 2} = 90^\circ
               \Rightarrow  {{B + C} \over 2} = 90^\circ - {A \over 2}
              sin\left( {{{B + C} \over 2}} \right) = 90^\circ - {A \over 2}
              \sin \left( {{{B + C} \over 2}} \right) = \cos {A \over 2} [Since,\sin \left( {90^\circ - \theta } \right) = \cos \theta ]


Q.7     Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Sol.       sin 67° + cos 75° = sin (90° – 23) + cos (90° – 15°)
                                           = cos 23° + sin 15°



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