If sec A-x-1-x-prove that A -tan A-2 x or 1-2 x-

Sec^2 A = 1 + tan^2 A x^2 + 1/2 + 1/(16x^2) = 1 + tan^2 A x^2 1/2 + 1/(16x^2) = tan^2 A ± [x 1/(4x)] = tan A sec A + tan A = x + 1/(4x)± (x 1/(4x)) = ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

If sec A=x+1/...

Question

If secA = $x + \frac{1}{x}$ ,prove that secA×tanA=2x or $\frac{1}{2}$ ×.

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Solution

$s e c^{2} A = 1 + t a n^{2} A x^{2} + \frac{1}{2} + \frac{1}{(16 x^{2})} = 1 + t a n^{2} A x^{2} - \frac{1}{2} + \frac{1}{(16 x^{2})} = t a n^{2} A \pm [x - \frac{1}{(4 x)}] = t a n A s e c A + t a n A = x + \frac{1}{(4 x)} \pm (x - \frac{1}{(4 x)}) = 2 x o r \frac{1}{(2 x)}$

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If secA = x+1x,prove that secA×tanA=2x or 12×.

sec2A=1+tan2Ax2+12+1(16x2)=1+tan2Ax2−12+1(16x2)=tan2A±[x−1(4x)]=tanAsecA+tanA=x+1(4x)±(x−1(4x))=2xor1(2x)

If secA = $x + \frac{1}{x}$ ,prove that secA×tanA=2x or $\frac{1}{2}$ ×.

$s e c^{2} A = 1 + t a n^{2} A x^{2} + \frac{1}{2} + \frac{1}{(16 x^{2})} = 1 + t a n^{2} A x^{2} - \frac{1}{2} + \frac{1}{(16 x^{2})} = t a n^{2} A \pm [x - \frac{1}{(4 x)}] = t a n A s e c A + t a n A = x + \frac{1}{(4 x)} \pm (x - \frac{1}{(4 x)}) = 2 x o r \frac{1}{(2 x)}$