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Standard IX

Mathematics

Values of Trigonometric Ratios

sin251o-x+sin...

Question

${sin}^{2} (51^{o} - x) + {sin}^{2} (39^{o} + x)$ =

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Solution

The correct option is C $1$
$s i n^{2} (51 - x) + s i n^{2} (39 + x)$
$s i n^{2} (51 - x) + c o s^{2} [90 - (39 + x)]$
$s i n^{2} (51 - x) + c o s^{2} (51 - x)$
$= 1 [∵ s i n^{2} θ + c o s^{2} θ = 1] .$

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, then:

{tan}^{- 1} \frac{1 - x}{1 + x} = \frac{1}{2} {tan}^{- 1} x = 0, x > 0

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sin2(51o−x)+sin2(39o+x)=

The correct option is C 1sin2(51−x)+sin2(39+x)sin2(51−x)+cos2[90−(39+x)]sin2(51−x)+cos2(51−x)=1[∵sin2θ+cos2θ=1].

${sin}^{2} (51^{o} - x) + {sin}^{2} (39^{o} + x)$ =

The correct option is C $1$
$s i n^{2} (51 - x) + s i n^{2} (39 + x)$
$s i n^{2} (51 - x) + c o s^{2} [90 - (39 + x)]$
$s i n^{2} (51 - x) + c o s^{2} (51 - x)$
$= 1 [∵ s i n^{2} θ + c o s^{2} θ = 1] .$