1.

SecD=5/4. Evaluate1-tanD/1+tanD

Answer»

ANSWER

GIVEN:

TO FIND:

\frac{1 - \tan \: d}{1 + \tan \: d}

SOLUTION:

Formula

1 + \tan {}^{2} d \: = \sec {}^{2} d

Putting Sec d = 5/4

\implies \: \tan {}^{2} d \: = \sec {}^{2} d - 1 \\ \implies \: \tan {}^{2} d = (\frac{5}{4}) { }^{2} - 1 \\ \implies \: \tan {}^{2} d \: = \frac{25}{16} - 1 \\ \implies \: \tan {}^{2} d \: = \frac{9}{16} \\ \implies \: \tan d \: = \sqrt{ \frac{9}{16} } \\ \implies \: \tan d = \frac{3}{4}

Putting tan d = 3/4

= \frac{1 - \tan \: d}{1 + \tan \: d} \\ = \frac{1 - \frac{3}{4} }{1 + \frac{3}{4} } \\ = \frac{ \frac{1}{4} }{ \frac{7}{4} } \\ = \frac{1}{4} \times \frac{4}{7} \\ = \frac{1}{7}

So , the value of the above expression is 1/7

SOME IMPORTANT FORMULA

Tan A = perpendicular/ base

Sin A = perpendicular/ HYPOTENUSE

Cos A = Base / hypotenuse


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