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Venkat Raj Binnani
Asked: 2022-11-01T23:31:12+05:30 In:

The value of \(\frac{{\left[ {4co{s^4}32^\circ cose{c^2}58^\circ – sin63^\circ cos27^\circ – co{s^2}63^\circ \; + \;1} \right]}}{{\left[ { – cose{c^2}24^\circ \; + \;ta{n^2}66^\circ – se{c^2}31^\circ \; + \;co{t^2}59^\circ } \right]{{\cos }^2}32}}\) is –
1. 1
2. -1
3. 2
4. -2

The value of \(\frac{{\left[ {4co{s^4}32^\circ cose{c^2}58^\circ – sin63^\circ cos27^\circ – co{s^2}63^\circ \; + \;1} \right]}}{{\left[ { – cose{c^2}24^\circ \; + \;ta{n^2}66^\circ – se{c^2}31^\circ \; + \;co{t^2}59^\circ } \right]{{\cos }^2}32}}\) is –
1. 1
2. -1
3. 2
4. -2
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1 Answer

  1. Correct Answer – Option 4 : -2

    GIVEN:

    \(\frac{{\left[ {4co{s^4}32^\circ cose{c^2}58^\circ – sin63^\circ cos27^\circ – co{s^2}63^\circ \; + \;1} \right]}}{{\left[ { – cose{c^2}24^\circ \; + \;ta{n^2}66^\circ – se{c^2}31^\circ \; + \;co{t^2}59^\circ } \right]{{\cos }^2}32^\circ }}\)

    FORMULA USED:

    sin(90° – x) = cos x, cosec(90° – x) = sec x

    (1 – sin2x) = cos2x

    (1 – cos2x) = sin2x

    CALCULATION:

    \(\frac{{\left[ {4co{s^4}32^\circ cose{c^2}58^\circ – sin63^\circ cos27^\circ – co{s^2}63^\circ \; + \;1} \right]}}{{\left[ { – cose{c^2}24^\circ \; + \;ta{n^2}66^\circ – se{c^2}31^\circ \; + \;co{t^2}59^\circ } \right]\;{{\cos }^2}32^\circ }}\)

    = \(\frac{{\left[ {4co{s^4}32^\circ se{c^2}32^\circ – sin63^\circ sin63^\circ – co{s^2}63^\circ \; + \;1} \right]}}{{\left[ { – se{c^2}66^\circ \; + \;ta{n^2}66^\circ – cose{c^2}59^\circ \; + \;co{t^2}59^\circ } \right]\;{{\cos }^2}32^\circ }}\)

    = \(\frac{{\left[ {4co{s^2}32^\circ – {{\sin }^2}63^\circ – co{s^2}63^\circ \; + \;1} \right]}}{{\left[ { – se{c^2}66^\circ \; + \;ta{n^2}66^\circ – cose{c^2}59^\circ \; + \;co{t^2}59^\circ } \right]\;{{\cos }^2}32^\circ }}\)

    = \(\frac{{\left[ {4co{s^2}32^\circ } \right]}}{{\left[ { – 1 – 1} \right]{{\cos }^2}32^\circ }}\)

    = 4/(- 2)

    = – 2

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