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TG EAMCET 2 May 2025 Shift 1 Paper

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Question : 69 of 160

Marks: +1, -0

If

1^{\circ}=0.0175

radians, then the approximate value of

\sec 58^{\circ}

1.9899
1.8788
1.8511
1.9677

Solution:

Let

f(x)=\sec (x)

Since

58^{\circ}

is close to

60^{\circ}

and

\sec (60^{\circ}) =2

Also,

60^{\circ}=60 \times 0.0175

radians

=1.05

Now,

f(x)=\sec x

\Rightarrow f'(x)\;=\sec (x) \tan (x) \\ f'(60^{\circ})\;=\sec (60^{\circ}) \tan (60^{\circ}) \\;=2 \times \sqrt{3}=2\sqrt{3}

And, change in angle,

\Delta x\;=58^{\circ}-68^{\circ}=-2^{\circ}

\;=-2 \times 0.0175 \;\text{ radians }\;

\;=-0.035 \;\text{ radians }\;

Now,

\Delta y \approx f'(x) \cdot \Delta x

\;\; \approx 2\sqrt{3} \times (-0.035)

\;\; \approx 2 \times 1.732 \times (-0.035)

\;\; \approx -0.12124

So,

\sec (58^{\circ}) \approx f (60^{\circ}) + \Delta y

\Rightarrow \;\; 2-0.12124 \approx 1.87876

So, the approximate value of

\sec (58^{\circ})

is 1.8788 .

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