Question

sin^2 28+ sin^2 62 + tan^2 38-cot^2 52 +1/4 sec^2 30

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$sin^{2}28+sin^{2}62+tan^{2}38-cot^{2}52+\frac{1}{4} sec^{2}30=\frac{4}{3}$

Step-by-step explanation:

$sin^{2}28+sin^{2}62+tan^{2}38-cot^{2}52+\frac{1}{4} sec^{2}30$

$=sin^{2}28+sin^{2}(90-28)+tan^{2}38-cot^{2}(90-38)+\frac{1}{4} \times \left(\frac{2}{\sqrt{3}}\right)^{2}$

$=(sin^{2}28+cos^{2}28)+tan^{2}38-tan^{2}38+\frac{1}{4} \times \frac{4}{3}$

$=1+\frac{1}{3}$

$=\frac{3+1}{3}\\=\frac{4}{3}$

Therefore,

$sin^{2}28+sin^{2}62+tan^{2}38-cot^{2}52+\frac{1}{4} sec^{2}30=\frac{4}{3}$

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