Question

Prove that

$\sf tan61=\dfrac{cos16+sin16}{cos16-sin16}$

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Answer 1

To Prove :-

• $\sf tan61=\dfrac{cos16+sin16}{cos16-sin16}$

Proof :-

• Taking RHS :-

$\sf\mapsto \dfrac{cos 16^o +sin16^o}{cos16^o - sin16^o}$

• Dividing Numerator and Denominator by cos16°

$\sf\mapsto \dfrac{\dfrac{cos 16^o +sin16^o}{cos16^o}}{\dfrac{cos16^o - sin16^o}{cos16^o}} \\\\\sf\mapsto \dfrac{\dfrac{cos16^o}{cos16^o}+\dfrac{sin16^o}{cos16^o}}{\dfrac{cos16^o}{cos16^o}-\dfrac{sin16^o}{cos16^o}}\\\\\sf\mapsto \dfrac{1+tan16^o}{1-tan16°}$

• We know that tan45° = 1 :-

$\sf\mapsto \dfrac{tan45^o+tan16^o}{1-tan45^o . tan16^o }$

• So that :-

$\sf\mapsto tan ( 45^o + 16^o) \qquad \bigg\lgroup \red{\bf tan(A+B)=\dfrac{tanAtanB}{1-tanA.tanB} }\bigg\rgroup \\\\\sf\mapsto \pink{ tan61^{\circ} }$

= LHS

Hence Proved !

Answer 2

Given,

$\tan(61) = \frac{ \cos(16) + \sin(16) }{ \cos(16) - \sin(16) }$

To Proof,

$\tan(61) = \frac{ \cos(16) + \sin(16) }{ \cos(16) - \sin(16) }$

Proof,

$\tan(61) = \frac{ \cos(16) + \sin(16) }{ \cos(16) - \sin(16) } \\ \\ \tan(61) = \frac{ \frac{\cos(16) + \sin(16) }{ \cos(16) } }{ \frac{\cos(16) - \sin(16) }{ \cos(16) } } \\ \\ \tan( 61 ) = \frac{ \frac{ \cos(16) }{ \cos(16) } + \frac{ \sin(16) }{ \cos(16) } }{ \frac{ \cos(16) }{ \cos(16) } - \frac{ \sin(16) }{ \cos(16) } } \\ \\ \tan(61) = \frac{1 + \tan(16) }{1 - \tan(16) }$

As tan(45) = 1

$\tan(61) = \frac{ \tan(45) + \tan(16) }{1 - 1 \times \tan(16) } \\ \\ \tan(61) = \frac{ \tan(45) + \tan(16) }{1 - \tan(45). \tan(16) }$

$As \: \: \tan(x + y) = \frac{ \tan(x) + \tan(y) }{1 - \tan(x) . \tan(y) }$

$\tan(61) = \frac{ \tan(45) + \tan(16) }{1 - \tan(45). \tan(16) } \\ \\ \tan(61) = \tan(45 + 16) \\ \\ \tan(61) = \tan(61)$

Hence Proved,