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Home » Prove that: (i) (cos 11o + sin 11o) / (cos 11o – sin 11o) = tan 56o (ii) (cos 9o + sin 9o) / (cos 9o – sin 9o) = tan 54o
Prove that: (i) (cos 11o + sin 11o) / (cos 11o – sin 11o) = tan 56o (ii) (cos 9o + sin 9o) / (cos 9o – sin 9o) = tan 54o
CBSE | Class 11 | Exercise 7.1 | Maths | RD Sharma | Values of Trigonometric Functions at Sum or Difference of Angles
Prove that: (i) (cos 11o + sin 11o) / (cos 11o – sin 11o) = tan 56o (ii) (cos 9o + sin 9o) / (cos 9o – sin 9o) = tan 54o

    \[\left(\mathbf{i} \right)~(cos\text{}{{11}^{o}}~+\text{}sin\text{}{{11}^{o}})\text{}/\text{}(cos\text{}{{11}^{o}}-\text{}sin\text{}{{11}^{o}})\text{}=\text{}tan\text{}{{56}^{o}}\]

  

Taking L.H.S:

    \[(cos\text{ }{{11}^{o}}~+\text{ }sin\text{ }{{11}^{o}})\text{ }/\text{ }(cos\text{ }{{11}^{o}}-\text{ }sin\text{ }{{11}^{o}})\]

Dividing numerator and denominator with

    \[cos\text{ }{{11}^{o}}~\]

, we get

    \[(cos\text{ }{{11}^{o}}~+\text{ }sin\text{ }{{11}^{o}})\text{ }/\text{ }(cos\text{ }{{11}^{o}}-\text{ }sin\text{ }{{11}^{o}})\]

    \[=\text{ }(1\text{ }+\text{ }tan\text{ }{{11}^{o}})\text{ }/\text{ }(1\text{ }-\text{ }tan\text{ }{{11}^{o}})\]

    \[=\text{ }(1\text{ }+\text{ }tan\text{ }{{11}^{o}})\text{ }/\text{ }(1-\text{ }1\times tan\text{ }{{11}^{o}})\]

or,

    \[=\text{ }(tan\text{ }{{45}^{o}}~+\text{ }tan\text{ }{{11}^{o}})\text{ }/\text{ }(1\text{ }\text{ }tan\text{ }{{45}^{o~}}\times \text{ }tan\text{ }{{11}^{o}})\]

Because,

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

So, we get

    \[(tan\text{ }{{45}^{o}}~+\text{ }tan\text{ }{{11}^{o}})\text{ }/\text{ }(1\text{ }-\text{ }tan\text{ }{{45}^{o~}}\times \text{ }tan\text{ }{{11}^{o}})\]

    \[=\text{ }tan\text{ }({{45}^{o}}~+\text{ }{{11}^{o}})\]

or,

    \[=\text{ }tan\text{ }{{56}^{o}}\]

    \[=\text{ }RHS\]

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

Hence Proved.

 

    \[\left( \mathbf{ii} \right)~(cos\text{ }{{9}^{o}}~+\text{ }sin\text{ }{{9}^{o}})\text{ }/\text{ }(cos\text{ }{{9}^{o}}-\text{ }sin\text{ }{{9}^{o}})\text{ }=\text{ }tan\text{ }{{54}^{o}}\]

Taking L.H.S, we get:

    \[(cos\text{ }{{9}^{o}}~+\text{ }sin\text{ }{{9}^{o}})\text{ }/\text{ }(cos\text{ }{{9}^{o}}-\text{ }sin\text{ }{{9}^{o}})\]

On Dividing numerator and denominator with

    \[cos\text{ }{{9}^{o}}~,\]

we get,

    \[(cos\text{ }{{9}^{o}}~+\text{ }sin\text{ }{{9}^{o}})\text{ }/\text{ }(cos\text{ }{{9}^{o}}-\text{ }sin\text{}{{9}^{o}})\]

or,

    \[=\text{ }(1\text{ }+\text{ }tan\text{ }{{9}^{o}})\text{ }/\text{ }(1\text{ }-\text{ }tan\text{ }{{9}^{o}})\]

    \[=\text{ }(1\text{ }+\text{ }tan\text{ }{{9}^{o}})\text{}/\text{}(1\text{ }-\text{ }1\text{ }\times \text{}tan\text{}{{9}^{o}})\]

or,

    \[=\text{ }(tan\text{}{{45}^{o}}~+\text{}tan\text{}{{9}^{o}})\text{}/\text{}(1\text{}-\text{}tan\text{}{{45}^{o}}~\times\text{}tan\text{}{{9}^{o}})\]

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

So, we get

    \[(tan{{45}^{o}}+tan{{9}^{o}})/(1-tan{{45}^{o}}~\times\text{ }tan\text{}{{9}^{o}})=tan({{45}^{o}}+{{9}^{o}})\]

or,

    \[=\text{ }tan\text{ }{{54}^{o}}\]

    \[=\text{ }RHS\]

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

Hence Proved.

i

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