Introduction To Trigonometry Exercise 8.4



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Q.1     Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Sol.       Consider a \Delta ABC, in which \angle B = 90^\circ .
              For \angle A, we have
              Base = AB, Perp = BC
              and Hyp = AC
              Therefore, \cot A = {{Base} \over {Perp}} = {{AB} \over {BC}}

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               \Rightarrow {{AB} \over {BC}} = \cot A = {{\cot A} \over 1}
              Let AB = k cotA and BC = k.
              By Pythagoras Theorem,
              AC = \sqrt {A{B^2} + B{C^2}} = \sqrt {{k^2}{{\cot }^2}A + {k^2}}
                     = k\sqrt {1 + {{\cot }^2}A}
              Therefore, \sin A = {{Perp} \over {Base}} = {{BC} \over {AC}} = {k \over {k\sqrt {1 + {{\cot }^2}A} }}
                                          = {1 \over {\sqrt {1 + {{\cot }^2}A} }}
              \sec A = {{Hyp} \over {Base}} = {{AC} \over {AB}} = {{k\sqrt {1 + {{\cot }^2}A} } \over {k\cot A}}
                         = \sqrt {{{1 + {{\cot }^2}A} \over {\cot A}}}

              and, \tan A = {{Perp} \over {Base}} = {{BC} \over {AB}} = {k \over {k\cot A}} = {1 \over {\cot A}}


 Q.2     Write the other trigonometric ratios of A in terms of secA.
Sol.        Consider a \Delta ABC, in which \angle B = 90^\circ .
               For \angle A, we have
               Base = AB, Perp = BC
               and Hyp = AC
               Therefore, \sec A = {{Hyp} \over {Base}} = {{AC} \over {AB}}
                \Rightarrow {{AC} \over {AB}} = \sec A = {{\sec A} \over 1}

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                \Rightarrow {{AB} \over {AC}} = {1 \over {\sec A}} = {{{1 \over {\sec A}}} \over 1} (Note this step)
               Let AB = k\left( {{1 \over {\sec A}}} \right),AC = k(1)
               By phythagoras Theorem
               BC = \sqrt {A{C^2} + A{B^2}} = \sqrt {{k^2} - {k^2}\left( {{1 \over {{{\sec }^2}A}}} \right)}
                      = k\sqrt {{{{{\sec }^2}A - 1} \over {{{\sec }^2}A}}} = {{k\sqrt {{{\sec }^2}A - 1} } \over {{{\sec }^2}A}}

               Therefore, \sin A = {{BC} \over {AC}} = {{k\sqrt {{{\sec }^2}A - 1} } \over {{{\sec A} \over k}}} = {{\sqrt {{{\sec }^2}A - 1} } \over {\sec A}}
               \cos A = {{AB} \over {AC}}{{k\left( {{1 \over {\sec A}}} \right)} \over k} = {1 \over {\sec A}}
               \tan A = {{BC} \over {AB}} = {{{{k\sqrt {{{\sec }^2}A - 1} } \over {\sec A}}} \over {k\left( {{1 \over {\sec A}}} \right)}} = \sqrt {{{\sec }^2}A - 1}
               \cot A = {1 \over {\tan A}} = {1 \over {\sqrt {{{\sec }^2}A - 1} }}
               \cos ecA = {1 \over {\sin A}} = {{\sec A} \over {\sqrt {{{\sec }^2}A - 1} }}


Q.3     Evaluate :
              (i) {{{{\sin }^2}63^\circ + {{\sin }^2}27^\circ } \over {{{\cos }^2}17^\circ + {{\cos }^2}73^\circ }}
            (ii) sin 25° cos 65° + cos25° sin65° [Since,sin\left( {90^\circ - \theta } \right) = \cos \theta ]
 Sol.       (i) Here, sin63° = sin(90° – 27°) = cos27°
               and cos17° = cos(90° – 73°) = sin 73° [Since,\cos \left( {90^\circ - \theta } \right) = \sin \theta ]
               Therefore, {{{{\sin }^2}63^\circ + {{\sin }^2}27^\circ} \over {{{\cos }^2}17^\circ + {{\cos }^2}73^\circ}} = {{{{\cos }^2}27^\circ + {{\sin }^2}27^\circ }\over {{{\sin }^2}73^\circ + {{\cos }^2}73^\circ}}
                 = {1 \over 1} = 1
                [Since,{\cos ^2}A + {\sin ^2}A = 1]

               (ii) sin25°cos65°+cos25°sin65°
               = sin(90°– 65°). cos65° + cos(90°– 65°) sin65°
               = cos65° cos65° + sin 65° sin65°
               [Since,\sin \left( {90^\circ - \theta } \right) = \cos \theta ]
              = {\cos ^2}65^\circ + {\sin ^2}65^\circ = 1 [Since,\cos \left( {90^\circ - \theta } \right) = \sin \theta ]

Q.4     Choose the correct option. Justify your choice :
             (i) 9{\sec ^2}A - 9{\tan ^2}A =
           (A) 1                                     (B) 9
           (C) 8                                      (D) 0

           (ii) (1+  tan\theta  + sec\theta )(1 + cos \theta – cosec \theta ) =
            (A) 0                                     (B) 1
            (C) 2                                     (D) None of these

            (iii) (secA + tanA)(1 – sinA) =
            (A) secA                              (B) sinA
            (C) cosecA                           (D) cosA

            (iv) {{1 + {{\tan }^2}A} \over {1 + {{\cot }^2}A}} =
            (A) {{{\sec }^2}A}                              (B) –1
            (C) {{{\cot }^2}A}                               (D) none of these
Sol.       (i) (B), because
              9{\sec ^2}A - 9{\tan ^2}A = 9\left( {{{\sec }^2}A - {{\tan }^2}A} \right)
               = 9 × 1 = 9 \left[ {Since,1 + {{\tan }^2}A = {{\sec }^2}A} \right]

              (ii) (C), because
              \left( {1 + \tan \theta + \sec \theta } \right)\left( {1 + \cot \theta - \cos ec\theta } \right)
               = \left( {1 + {{\sin \theta } \over {\cos \theta }} + {1 \over {\cos \theta }}} \right)\left( {1 + {{\cos \theta } \over {\sin \theta }} - {1 \over {\sin \theta }}} \right)
               = \left( {{{\cos \theta + \sin \theta + 1} \over {\cos \theta }}} \right)\left( {{{\sin \theta + \cos \theta - 1} \over {\sin \theta }}} \right)
              = {{{{\left( {\cos \theta + \sin \theta} \right)}^2} - 1} \over {\sin \theta \cos \theta }} [Since,\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^2}]
              =  {{\left( {{{\cos }^2}\theta + {{\sin }^2}\theta} \right) - 2\cos \theta \sin \theta - 1} \over {\sin \theta \cos \theta }} [Since,{\sin ^2}\theta + {\cos ^2}\theta = 1]
              = {{1 + 2\cos \theta \sin \theta - 1} \over {\sin \theta \cos \theta }} = {{2\cos \theta \sin \theta } \over {\sin \theta \cos \theta }} = 2

             (iii) (D), because
             (secA + tanA) (1 – sinA)
             = \left( {{1 \over {\cos A}} + {{\sin A} \over {\cos A}}} \right)\left( {1 - \sin A} \right)
             = \left( {{{1 + \sin A} \over {\cos A}}} \right)\left( {1 - \sin A} \right) [Since,\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^2}]
               = {{1 - {{\sin }^2}A} \over {\cos A}}{{{{\cos }^2}A} \over {\cos A}} = \cos A[Since,{\sin ^2}A + {\cos ^2}A = 1]

              (iv) (D), because
              {{1 + {{\tan }^2}A} \over {1 - {{\cot }^2}A}} = {{1 + {{\tan }^2}A} \over {1 + {1 \over {{{\tan }^2}A}}}} = {{1 + {{\tan }^2}A} \over {{{{{\tan }^2}A + 1} \over {{{\tan }^2}A}}}}
                = \left( {1 + {{\tan }^2}A} \right) \times {{{{\tan }^2}A} \over {1 + {{\tan }^2}A}} = {\tan ^2}A


Q.5     Prove the following identities, where the angles involved are acute angles for which the expressions are defined :
           (i) {\left( {\cos ec\theta - \cot \theta } \right)^2} = {{1 - \cos \theta } \over {1 + \cos \theta }}
           (ii) {{\cos A} \over {1 + \sin A}} + {{1 + \sin A} \over {\cos A}} = 2\sec A
           (iii) {{\tan \theta } \over {1 - \cot \theta }} + {{\cot \theta } \over {1 - \tan \theta }} = 1 + \sec \theta \cos ec\theta
           (iv) {{1 + \sec A} \over {\sec A}} = {{{{\sin }^2}A} \over {1 - \cos A}}
           (v) {{\cos A - \sin A+ 1} \over {\cos A + \sin A - 1}} = \cos ecA + \cot A, using the identity \cos e{c^2}A = 1 + {\cot ^2}A
           (vi) \sqrt {{{1 + \sin A} \over {1 - \sin A}} = } \sec A + \tan A
           (vii) {{\sin \theta - 2{{\sin }^3}\theta } \over {2{{\cos }^3}\theta - \cos \theta }} = \tan \theta
           (viii) {{\rm{(sinA + cosecA)}}^2} + {{\rm{(cosA + secA)}}^2} = 7 + {\tan ^2}A + {\cot ^2}A
           (ix) (cosecA – sinA)(secA – cosA) = {1 \over {\tan A + \cot A}}
           (x) \left( {{{1 + {{\tan }^2}A} \over {1 + {{\cot }^2}A}}} \right) = {\left( {{{1 - \tan A} \over {1 - {{\cot }^2}A}}} \right)^2} = {\tan ^2}A
Sol.       (i) We have,
              L.H.S. = {{\rm{(cosec}}\theta - \cot \theta {\rm{ )}}^2}
               = {\left( {{1 \over {\sin \theta }} - {{\cos \theta } \over {\sin \theta }}} \right)^2} = {\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right)^2}
               = {{{{\left( {1 - \cos \theta } \right)}^2}} \over {{{\sin }^2}\theta }} = {{{{\left( {1 - \cos \theta } \right)}^2}} \over {1 - {{\cos }^2}\theta }} [Since,{\sin ^2}\theta = 1 - {\cos ^2}\theta ]
               = {{{{\left( {1 - \cos \theta } \right)}^2}} \over {\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)}} = {{1 - \cos \theta } \over {1 + \cos \theta }}
              R.H.S. [Since,{A^2} - {B^2} = \left( {A + B} \right)\left( {A - B} \right)]

              (ii) We have,
              L.H.S. = {{\cos A} \over {1 + \sin A}} + {{1 + \sin A} \over {\cos A}}
              = {{{{\cos }^2}A + {{\left( {1 + \sin A} \right)}^2}} \over {\cos A\left( {1 + \sin A} \right)}}
              = {{{{\cos }^2}A + 1 + 2\sin A + {{\sin }^2}A} \over {\cos A\left( {1 + \sin A} \right)}}
              = {{\left( {{{\cos }^2}A + {{\sin }^2}A} \right) + 1 + 2\sin A} \over {\cos A\left( {1 + \sin A} \right)}}
              = {{1 + 1 + 2\sin A} \over {\cos A\left( {1 + \sin A} \right)}}[Since {\sin ^2}A + {\cos ^2}A = 1]
              = {{2 + 2\sin A} \over {\cos A\left( {1 + \sin A} \right)}} = {{2\left( {1 + \sin A} \right)} \over {\cos A\left( {1 + \sin A} \right)}}
              = {2 \over {\cos A}} = 2\sec A = R.H.S.

               (iii) We have,
               L.H.S. = {{\tan \theta } \over {1 - \cot \theta }} + {{\cot \theta } \over {1 - \tan \theta }}
               = {{\tan \theta } \over {1 - {1 \over {\tan \theta }}}} + {{{1 \over {\tan \theta }}} \over {1 - \tan \theta }}
                = {{\tan \theta } \over {{{\tan \theta - 1} \over {\tan \theta }}}} + {1 \over {\tan \theta \left( {1 - \tan \theta } \right)}}
               
 = {{{{\tan }^2}\theta } \over {\tan \theta - 1}} + {1 \over {\tan \theta \left( {1 - \tan \theta } \right)}}
                = {{{{\tan }^2}\theta } \over {\tan \theta - 1}} - {1 \over {\tan \theta \left( {\tan \theta - 1} \right)}}
               = {{{{\tan }^3}\theta - 1} \over {\tan \theta \left( {\tan \theta - 1} \right)}}
               = {{\left( {\tan \theta - 1} \right)\left( {{{\tan }^2}\theta + \tan \theta + 1} \right)} \over {{{\tan }}\theta \left( {\tan \theta - 1} \right)}}
               [Since,{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)]
                = {{{{\tan }^2}\theta + \tan \theta + 1} \over {{{\tan }^2}\theta }}
                = {{{{\tan }^2}\theta } \over {\tan \theta }} + {{\tan \theta } \over {\tan \theta }} + {1 \over {\tan \theta }}
                = \tan \theta + 1 + \cot \theta = 1 + \tan \theta + \cot \theta
                = 1 + {{\sin \theta } \over {\cos \theta }} + {{\cos \theta } \over {\sin \theta }}
                = 1 + {{{{\sin }^2}\theta + {{\cos }^2}\theta } \over {\cos \theta }}
                = 1 + {1 \over {\sin \theta \cos \theta }} = 1 + \cos ec\theta \sec \theta
                = R.H.S.

                 (iv) R.H.S. = {{{{\sin }^2}A} \over {1 - \cos A}} = {{1 - {{\cos }^2}A} \over {1 - \cos A}}
                 [Since,{\sin ^2}A = 1 - {\cos ^2}A]
                 = {{\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)} \over {1 - \cos A}} = 1 + \cos A
                 [Since,{A^2} - {B^2} = \left( {A + B} \right)\left( {A - B} \right)]
                 = 1 + {1 \over {\sec A}} = {{1 + \sec A} \over {\sec A}} = L.H.S.

                 (v) L.H.S. = {{\cos A - \sin A + 1} \over {\cos A + \sin A - 1}} = {{{{\cos A - \sin A + 1} \over {\sin A}}} \over {{{\cos A + \sin A - 1} \over {\sin A}}}}
                 = {{\cos A - 1 + \cos ecA} \over {\cos A + 1 - \cos ecA}}
                 [Since,1 + {\cot ^2}A = \cos e{c^2}A]
                  = {{\cot A + \cos ecA - \left( {\cos e{c^2}A - {{\cot }^2}A} \right)} \over {\cot A - \cos ecA + 1}}
                 = {{\cot A + \cos ecA - \left( {\cos ecA + \cot A} \right)\left( {\cos ecA - \cot A} \right)} \over {\cot A - \cos ecA + 1}}
                  [Since{A^2} - {B^2} = \left( {A + B} \right)\left( {A - B} \right)]
                  Taking common(cosecA + cotA)
                  = {{\left( {\cos ecA + \cot A} \right)\left( {1 - \cos ecA + \cot } \right)} \over {\left( {\cot A - \cos ecA + 1} \right)}}
                   = cosec A + cot A
                   = R.H.S.

                  (vi) We have,
                  L.H.S. = \sqrt {{{1 + \sin A} \over {1 - \sin A}}} = \sqrt {{{1 + \sin A} \over {1 - \sin A}} \times {{1 + \sin A} \over {1 + \sin A}}}
                  [Multiplying and dividing by ] \sqrt {1 + \sin A}
                  = \sqrt {{{{{\left( {1 + \sin A} \right)}^2}} \over {1 - {{\sin }^2}A}}} = \sqrt {{{{{\left( {1 + \sin A} \right)}^2}} \over {{{\cos }^2}A}}}  [Since,\,{\sin ^2}A + {\cos ^2}A = 1]
                   = {\sqrt {\left( {{{1 + \sin A} \over {\cos A}}} \right)} ^2} = {{1 + \sin A} \over {\cos A}}
                   = {1 \over {\cos A}} + {{\sin A} \over {\cos A}} = \sec A + \tan A
                   = R.H.S. \left[ {Since,\,\tan A = {{\sin A} \over {\cos A}}} \right]

                   (vii) We, have,
                   L.H.S. = {{\sin \theta - 2{{\sin }^3}\theta } \over {2{{\cos }^3}\theta - \cos \theta }} = {{\sin \theta (1 - 2{{\sin }^2}\theta )} \over {\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}
                    = \tan \theta \left[ {{{1 - 2{{\sin }^2}\theta } \over {2\left( {1 - {{\sin }^2}\theta } \right) - 1}}} \right]
                    = \tan \theta \left[ {{{1 - 2{{\sin }^2}\theta } \over {2 - 2{{\sin }^2}\theta - 1}}} \right]
                     = \tan \theta \left[ {{{1 - 2{{\sin }^2}\theta } \over {1 - 2{{\sin }^2}\theta }}} \right] = \tan \theta \times 1
                     = \tan \theta = R.H.S.

                     (viii) We have,
                     L.H.S. = {\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \sec A} \right)^2}
                     = ({\sin ^2}A + \cos e{c^2}A + 2\sin A\cos ecA) + \left( {{{\cos }^2}A + {{\sec }^2}A + 2\cos A\sec A} \right)
                      = \left( {{{\sin }^2}A + \cos e{c^2}A + 2\sin A.{1 \over {\sin A}}} \right) + \left( {{{\cos }^2}A + {{\sec }^2}A + 2\cos A.{1 \over {\cos A}}} \right)
                      = \left( {{{\sin }^2}A + \cos e{c^2}A + 2} \right) + \left( {{{\cos }^2}A + {{\sec }^2}A + 2} \right)
                      = {\sin ^2}A + {\cos ^2}A + \cos e{c^2}A + {\sec ^2}A + 4 [Since,\,\,{\sin ^2}A + {\cos ^2}A = 1]
                      = 1 + \left( {1 + {{\cot }^2}A} \right) + \left( {1 + {{\tan }^2}A} \right) + 4
                      = 7 + {\tan ^2}A + {\cot ^2}A
                      = [Since,\,\cos e{c^2}A = 1 + {\cot ^2}A\,\,and\,{\sec ^2}A = 1\, + \,{\tan ^2}A]
                      = R.H.S.

                      (ix) We have,
                      L.H.S. = (cosec A – sinA) (secA – cosA)
                      = \left( {{1 \over {\sin A}} - \sin A} \right)\left( {{1 \over {\cos A}} - \cos A} \right)
                      = \left( {{{1 - {{\sin }^2}A} \over {\sin A}}} \right)\left( {{{1 - {{\cos }^2}A} \over {\cos A}}} \right)
                      = {{{{\cos }^2}A} \over {\sin A}} \times {{{{\sin }^2}A} \over {\cos A}}
                      = sinA cosA
                      = {{\sin A\cos A} \over {{{\sin }^2}A + {{\cos }^2}A}} [Since,{\sin ^2}A + {\cos ^2}A = 1]
                      Dividing Numerator and Denominator by sinA cosA
                      {{{{\sin A\cos A} \over {\sin A\cos A}}} \over {{{{{\sin }^2}A} \over {\sin A\cos A}} + {{{{\cos }^2}A} \over {\sin A\cos A}}}}
                      = {1 \over {{{\sin A} \over {\cos A}} + {{\cos A} \over {\sin A}}}}
                      = {1 \over {\tan A + \cot A}} = R.H.S.

                       (x) We have,
                       L.H.S. = \left( {{{1 + {{\tan }^2}A} \over {1 + {{\cot }^2}A}}} \right) = {{{{\sec }^2}A} \over {\cos e{c^2}A}}
                       = {1 \over {{{\cos }^2}A}} \times {{{{\sin }^2}A} \over 1} = {\tan ^2}A
                       R.H.S. = {\left( {{{1 - \tan A} \over {1 - \cot A}}} \right)^2} = {\left( {{{1 - \tan A} \over {1 - {1 \over {\tan A}}}}} \right)^2}
                       = {\left( {{{1 - \tan A} \over {{{\tan A - 1} \over {\tan A}}}}} \right)^2} = {\left( { - \tan A} \right)^2} = {\tan ^2}A
                        Therefore, L.H.S. = R.H.S.



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