Question

cosec2 (90°- theta) - tan2 theta/ 2 (cos2 48° + cos2 42°) - 2tan2 30° sec2 52° sin2 38°/cosec2 70°- tan2 20°

Answer 1

Solution:

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We are using following :

i) $cosec(90-\theta) = sec\theta$

ii) We know the Trigonometric identity:

Sec²A - tan²A = 1

iii) $cos48=cos(90-42) = sin42$

iv) sin²A + cos²A = 1

v) tan30° = 1/√3

vi) sec52° = sec(90-38) =cosec38°

vii ) cosecAsinA = 1

viii)cosec70= cosec(90-20)

= sec20°

ix ) cosec²A - cot²A = 1

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Now ,

LHS =

$\frac{cosec^{2}(90-\theta)-tan^{2}\theta}{2(cos^{2}48+cos^{2}42)}-\frac{2tan^{2}30sec^{2}52sin^{2}38}{cosec^{2}70-tan^{2}20}$

= $\frac{sec^{2}\theta-tan^{2}\theta}{2(sin^{2}42+cos^{2}42)}-\frac{2\times\left(\frac{1}{\sqrt{3}}\right)^{2}cosec^{2}38sin^{2}38}{sec^{2}20-tan^{2}20}$

= $\frac{1}{2\times 1}-\frac{2\times\frac{1}{3}\times 1}{1}$

= $\frac{1}{2}-\frac{2}{3}$

= $\frac{(3-4)}{6}$

= $\frac{-1}{6}$

Therefore,

$\frac{cosec^{2}(90-\theta)-tan^{2}\theta}{2(cos^{2}48+cos^{2}42)}-\frac{2tan^{2}30sec^{2}52sin^{2}38}{cosec^{2}70-tan^{2}20}$ = -1/6

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Answer 2

Step-by-step explanation:

here......

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