Math, asked by srinethi15, 1 year ago

the values of tan theta from the equ 3 (sec square theta -1)+16tan theta+5=0are

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Answered by waqarsd
2
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Answered by ashpatel38
0
3(sec^2 x - 1) + 16tanx + 5 = 0

We know, 1 + tan^2 x = sec^2 x

3tan^2 x + 16tanx + 5= 0

Let tanx = a
3a^2 + 16a + 5 = 0
3a^2 + 15a + a + 5 = 0
(a+5) (3a+1) = 0

a = -5, -1/3

=> -5 does not come in the range of the value of function tanx

=> tanx = -1/3

Hope this helps... plsss make it the brainliest
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