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Introduction to Trigonometry : Exercise - 8.2 (Mathematics NCERT Class 10th)


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Q.1     Evaluate :
           (i) sin 60 ° cos 30° + sin 30° cos 60°
           (ii) 2{\tan ^2}45^\circ + {\cos ^2}30^\circ - {\sin ^2}60^\circ  
           (iii) {{\cos 45^\circ } \over {\sec 30^\circ + \cos ec30^\circ }}
           (iv) {{\sin 30^\circ + \tan 45^\circ - \cos ec60^\circ } \over {\sec 30^\circ + \cos 60^\circ + \cot 45^\circ }}
            (v) {{5{{\cos }^2}60^\circ + 4{{\sec }^2}30^\circ - {{\tan }^2}45^\circ } \over {{{\sin }^2}30^\circ + {{\cos }^2}30^\circ }}
Sol.     (i) sin 60° cos 30° + sin 30° cos 60°
             = {{\sqrt 3 } \over 2} \times {{\sqrt 3 } \over 2} + {1 \over 2} \times {1 \over 2}

             = {3 \over 4} + {1 \over 4} = {{3 + 1} \over 4} = {4 \over 4} = 1

            (ii) 2{\tan ^2}45^\circ + {\cos ^2}30^\circ - {\sin ^2}60^\circ
             = 2{\left( 1 \right)^2} + {\left( {{{\sqrt 3 } \over 2}} \right)^2} - {\left( {{{\sqrt 3 } \over 2}} \right)^2}
             = 2 + {3 \over 4} - {3 \over 4} = 2

            (iii) {{\cos 45^\circ } \over {\sin 30^\circ + \cos ec30^\circ }}
              = {{{1 \over {\sqrt 2 }}} \over {{2 \over {\sqrt 3 }} + 2}} = {{{1 \over {\sqrt 2 }}} \over {2 + {{2\sqrt 3 } \over {\sqrt 3 }}}} = {1 \over {\sqrt 2 }} \times {{\sqrt 3 } \over {2 + 2\sqrt 3 }}
             = {{\sqrt 3 } \over {\sqrt 2 \times 2\left( {\sqrt 3 + 1} \right)}} \times {{\sqrt 3 - 1} \over {\sqrt 3 - 1}}
             = {{\sqrt 3 \left( {\sqrt 3 - 1} \right)} \over {\sqrt 2 \times 2\left( {3 - 1} \right)}} = {{\sqrt 2 \times \sqrt 3 \left( {\sqrt 3 - 1} \right)} \over {\sqrt 2 \times \sqrt 2 \times 2 \times 2}}
             = {{3\sqrt 2 - \sqrt 6 } \over 8}

             (iv) {{\sin 30^\circ + \tan 45^\circ - \cos ec60^\circ } \over {\sin 30^\circ + \cos 45^\circ + \cot 45^\circ }}
             {{{1 \over 2} + 1 - {2 \over {\sqrt 3 }}} \over {{2 \over {\sqrt 3 }} + {1 \over 2}}} = {{{{1 + 2} \over 2} - {2 \over {\sqrt 3 }}} \over {{2 \over {\sqrt 3 }} + {{1 + 2} \over 2}}}{\mkern 1mu}
               = {{{3 \over 2} - {2 \over {\sqrt 3 }}} \over {{2 \over {\sqrt 3 }} + {3 \over 2}}} = {{{{3\sqrt 3 - 4} \over {2\sqrt 3 }}} \over {{{4 + 3\sqrt 3 } \over {2\sqrt 3 }}}}
               = {{3\sqrt 3 - 4} \over {4 + 3\sqrt 3 }} = {{\left( {3\sqrt 3 - 4} \right)\left( {3\sqrt 3 - 4} \right)} \over {\left( {4 + 3\sqrt 3 } \right)\left( {3\sqrt 3 - 4} \right)}}
              {{27 + 16 - 12\sqrt 3 - 12\sqrt 3 } \over {27 - 16}}
              {{43 - 24\sqrt 3 } \over {11}}

              (v) {{5{{\cos }^2}60^\circ + 4{{\sec }^2}30^\circ - {{\tan }^2}45^\circ } \over {{{\sin }^2}30^\circ + {{\cos }^2}30^\circ }}
Sol.       {{5{{\left( {{1 \over 2}} \right)}^2} + 4{{\left( {{2 \over {\sqrt 3 }}} \right)}^2} - {{\left( 1 \right)}^2}} \over {{{\left( {{1 \over 2}} \right)}^2} + {{\left( {{{\sqrt 3 } \over 2}} \right)}^2}}}
              {{5 \times {1 \over 4} + 4 \times {4 \over 3} - 1} \over {{1 \over 4} + {3 \over 4}}} = {5 \over 4} + {{16} \over 3}-1 = {{{1 \over {12}}\left( {15 + 64 - 12} \right)} \over {{{1 + 3} \over 4}}}
              {{{1 \over {12}} \times 67} \over {{4 \over 4}}} = {{67} \over {12}}


Q.2     Choose the correct option and justify :
           (i) {{2\tan 30^\circ } \over {1 + {{\tan }^2}30^\circ }} =
           (A) sin 60°                    (B) cos 60°
           (C) tan 60°                   (D) sin 30°

           (ii) {{1 - {{\tan }^2}45^\circ } \over {1 + {{\tan }^2}45^\circ }} =
           (A) tan 90°                   (B) 1
           (C) sin 45 °                   (D) 0

           (iii) sin 2A = 2 sinA is true when A =
            (A) 0°                           (B) 30°
            (C) 45°                         (D) 60°

            (iv) {{2{{\tan }^2}30^\circ } \over {1 - {{\tan }^2}30^\circ }} =
            (A) cos 60°                   (B) sin 60°
            (C) tan 60 °                  (D) none of these
Sol.        (i) (A)
               Because {{2\tan 30^\circ } \over {1 + {{\tan }^2}30^\circ }} = {{2 \times {1 \over {\sqrt 3 }}} \over {1 + {{\left( {{1 \over {\sqrt 3 }}} \right)}^2}}} = {{{2 \over {\sqrt 3 }}} \over {1 + {1 \over 3}}}
                = {2 \over {\sqrt 3 }} \times {3 \over {3 + 1}} = {2 \over {\sqrt 3 }} \times {3 \over 4}
                = {{\sqrt 3 } \over 2} = \sin 60^\circ       

               (ii) (D)
               Because {{1 - {{\tan }^2}45^\circ } \over {1 + {{\tan }^2}45^\circ }} = {{1 - 1} \over {1 + 1}} = {0 \over 2} = 0

               (iii) (A)
               Because when A = 0, sin 2 A = sin 0 = 0
               and, 2 sinA = 2 sin 0 = 2 × 0 = 0
                \Rightarrow sin 2A = 2sinA, when A = 0

               (iv) (C)
               Because {{2\tan 30^\circ } \over {1 - {{\tan }^2}30^\circ }} = {{2 \times {1 \over {\sqrt 3 }}} \over {1 - {{\left( {{1 \over {\sqrt 3 }}} \right)}^2}}} = {{{2 \over {\sqrt 3 }}} \over {1 - {1 \over 3}}}
                                                   = {2 \over {\sqrt 3 }} \times {3 \over {3 - 1}} = {2 \over {\sqrt 3 }} \times {3 \over 2}
                                                   = \sqrt 3 = \tan 60^\circ


Q.3     If tan (A + B) = \sqrt 3 and tan (A – B) = B," /> find A and B.
Sol.       tan (A + B) = {\sqrt 3 }
              \Rightarrow \tan \left( {A + B} \right) = \tan 60^\circ
              \Rightarrow A + B = 60^\circ ...(1)
             \tan \left( {A - B} \right) = {1 \over {\sqrt 3 }}
              \Rightarrow \tan \left( {A - B} \right) = \tan 30^\circ
              \Rightarrow A – B = 30 °...(2)
             Solving (1) and (2), we get
             A = 45° and B = 15°
             Hence, A = 45° and B = 15°


Q.4     State whether the following are true of false. Justify your answer.
           (i) sin (A + B) = sin A + sin B.
           (ii) The value of sin \theta increases as \theta increases.
           (iii) The value of cas\theta increases as \theta increases.
           (iv) sin\theta = cos\theta for all values of \theta .
           (v) cot A is not defined for A = 0°.
Sol.       (i) False Because
             When A = 60° and B = 30°. Then,
              sin (A + B) = sin (60° + 30°) = sin 90° = 1
              and, sin A + sin B = sin 60° + sin 30°
                                       = {{\sqrt 3 } \over 2} + {1 \over 2} = {{\sqrt 3 + 1} \over 2}
               \Rightarrow sin (A + B)  \ne sin A + sin B
              (ii) True Because
              Clearly from the table below :

10
             We find that the value of sin\theta increases as \theta increases.

             (iii) False Because
             Clearly from the table below :

11

              We find that the value of cos\theta decreases as \theta increases.

              (iv)  False as it is only true for \theta = 45°.
              \left( {\sin 45^\circ = {1 \over {\sqrt 2 }} = \cos 45^\circ } \right)

               (v) True Because tan 0° = 0 and \cot 0^\circ = {1 \over {\tan 0^\circ }} = {1 \over 0},i.e., undefined.



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