How do you simplify $[\frac{2 + {tan}^{2} x}{{sec}^{2} x}] - 1$ $[(2 + tan^2x) / sec^2x] - 1$ ?

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How do you simplify $[\frac{2 + {tan}^{2} x}{{sec}^{2} x}] - 1$ $[(2 + tan^2x) / sec^2x] - 1$ ?

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Answer 1 · 2018-04-19T17:40:53

${cos}^{2} x$ $cos^2x$ .

Explanation:

Since, ${sec}^{2} x = 1 + {tan}^{2} x$ $sec^2x=1+tan^2x$ , we have,

$[\frac{2 + {tan}^{2} x}{{sec}^{2} x}] - 1$ $[(2+tan^2x)/sec^2x]-1$ ,

$= [\frac{(2 + {tan}^{2} x) - {sec}^{2} x}{{sec}^{2} x}]$ $=[{(2+tan^2x)-sec^2x}/sec^2x]$ ,

$= [\frac{2 - ({sec}^{2} x - {tan}^{2} x)}{{sec}^{2} x}]$ $=[{2-(sec^2x-tan^2x)}/sec^2x]$ ,

$= \frac{2 - 1}{{sec}^{2} x}$ $=(2-1)/sec^2x$ ,

$= \frac{1}{{sec}^{2} x}$ $=1/sec^2x$ ,

$= {cos}^{2} x$ $=cos^2x$ .

Answer 2 · 2018-04-19T17:45:06

${cos}^{2} x$ $cos^2x$

Explanation:

$using the trigonometric identity$ $"using the "color(blue)"trigonometric identity"$

$∙ x 1 + {tan}^{2} x = {sec}^{2} x$ $•color(white)(x)1+tan^2x=sec^2x$

$\Rightarrow [\frac{1 + 1 + {tan}^{2} x}{{sec}^{2} x}] - \frac{{sec}^{2} x}{{sec}^{2} x}$ $rArr[(1+1+tan^2x)/sec^2x]-sec^2x/sec^2x$

$= \frac{1 + {sec}^{2} x}{{sec}^{2} x} - \frac{{sec}^{2} x}{{sec}^{2} x}$ $=(1+sec^2x)/sec^2x-sec^2x/sec^2x$

$= \frac{1}{{sec}^{2} x} = {cos}^{2} x$ $=1/sec^2x=cos^2x$

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How do you simplify $[\frac{2 + {tan}^{2} x}{{sec}^{2} x}] - 1$ $[(2 + tan^2x) / sec^2x] - 1$ ?

2 Answers

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