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Home » Prove that: (cos 8degree – sin 8degree) / (cos 8degree + sin 8degree) = tan 37degree
Prove that: (cos 8degree – sin 8degree) / (cos 8degree + sin 8degree) = tan 37degree
CBSE | Class 11 | Exercise 7.1 | Maths | RD Sharma | Values of Trigonometric Functions at Sum or Difference of Angles
Prove that: (cos 8degree – sin 8degree) / (cos 8degree + sin 8degree) = tan 37degree

    \[(cos\text{ }{{8}^{o}}-\text{ }sin\text{ }{{8}^{o}})\text{ }/\text{ }(cos\text{ }{{8}^{o}}~+\text{ }sin\text{ }{{8}^{o}})\text{ }=\text{ }tan\text{ }{{37}^{o}}\]

Let us consider LHS:

    \[(cos\text{ }{{8}^{o}}-\text{ }sin\text{ }{{8}^{o}})\text{ }/\text{ }(cos\text{ }{{8}^{o}}~+\text{ }sin\text{ }{{8}^{o}})\]

Now let us divide the numerator and denominator by

    \[cos\text{ }{{8}^{o}}~\]

    \[(cos{{8}^{o}}-sin{{8}^{o}})/(cos{{8}^{o}}+sin{{8}^{o}})=(1tan{{8}^{o}})/(1+tan{{8}^{o}})\]

    \[=\text{ }(1\text{ }-\text{ }tan\text{ }{{8}^{o}})\text{ }/\text{ }(1\text{ }+\text{ }1\times tan\text{ }{{8}^{o}})\]

    \[=\text{ }(tan\text{ }{{45}^{o}}-\text{ }tan\text{ }{{8}^{o}})\text{ }/\text{ }(1\text{ }+\text{ }tan\text{ }{{45}^{o}}~\times tan\text{ }{{8}^{o}})\]

Since,

    \[tan\text{}\left(A+B \right)\text{}=\text{}\left(tan\text{}A\text{}+\text{}tan\text{}B\right)\text{}/\text{}\left(1\text{}-\text{}tan\text{}A\text{}tan\text{}B \right)\]

Therefore,

    \[(tan{{45}^{o}}-tan{{8}^{o}})/(1+tan{{45}^{o}}-\timestan\text{}{{8}^{o}})=tan({{45}^{o}}-{{8}^{o}})\]

    \[=\text{ }tan\text{ }{{37}^{o}}\]

    \[=\text{ }RHS\]

    \[\therefore LHS\text{ }=\text{ }RHS\]

Hence proved.

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