If \(\frac{tan\,26°+tan\,19°}{x(1-tan\,26°tan\,19°)}\) = cos 60° then the value of x is
(a) 1
(b) √2
(c) 2
(d) √3
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If \(\frac{tan\,26°+tan\,19°}{x(1-tan\,26°tan\,19°)}\) = cos 60° then the value of x is
(a) 1
(b) √2
(c) 2
(d) √3
If \(\frac{tan\,26°+tan\,19°}{x(1-tan\,26°tan\,19°)}\) = cos 60° then the value of x is
(a) 1
(b) √2
(c) 2
(d) √3
(c) 2
\(\frac{tan\,26°+tan\,19°}{x(1-tan\,26°tan\,19°)}\) = cos 60°
= \(\frac{tan\,26°+tan\,19°}{1-tan\,26°tan\,19°}\) = x cos 60°
= tan (26° + 19°) = \(x\) x \(\frac12\) \(\bigg[∵ tan (A + B)=\frac{tan\,A+tan\,B}{1-tan\,A\,tan\,B}\bigg]\)
= tan 45° = \(\frac{x}{2}\) ⇒ \(\frac{x}{2}\) = 1 ⇒ x = 2.