If x-tan- and y-cosec- thenA- x y-1-x-yB- x y-1-x-yC- x y-1-x-2 yD- x y-1-y-x

The correct option is D xy+1 =y xxy+1=(sec ϕ tan ϕ)(cosec ϕ +cot ϕ)+1 =1 sin ϕ+cos ϕ/sin ϕ cos ϕ=sin^2 ϕ+cos^2 ϕ sin ϕ +cos ϕ/sin ϕ cos ϕ =(cosec ϕ+cot ϕ) ( ...

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Byju's Answer

Standard XII

Mathematics

Principal Solution of Trigonometric Equation

If x=ϕ-tanϕ a...

Question

If $x = s e c ϕ - t a n ϕ$ and $y = c o s e c ϕ + c o t ϕ$ , then

xy +1 =x -y

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xy+1 =x -2y

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xy +1 =x+y

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xy+1 =y-x

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Solution

The correct option is D xy+1 =y-x
$x y + 1 = (s e c ϕ - t a n ϕ) (c o s e c ϕ + c o t ϕ) + 1 = \frac{1 - s i n ϕ + c o s ϕ}{s i n ϕ c o s ϕ} = \frac{s i n^{2} ϕ + c o s^{2} ϕ - s i n ϕ + c o s ϕ}{s i n ϕ c o s ϕ} = (c o s e c ϕ + c o t ϕ) - (s e c ϕ - t a n ϕ) = y - x$

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If x=secϕ−tanϕ and y=cosecϕ+cotϕ, then

The correct option is D xy+1 =y-xxy+1=(secϕ−tanϕ)(cosecϕ+cotϕ)+1=1−sinϕ+cosϕsinϕcosϕ=sin2ϕ+cos2ϕ−sinϕ+cosϕsinϕcosϕ=(cosecϕ+cotϕ)−(secϕ−tanϕ)=y−x

If $x = s e c ϕ - t a n ϕ$ and $y = c o s e c ϕ + c o t ϕ$ , then

The correct option is D xy+1 =y-x
$x y + 1 = (s e c ϕ - t a n ϕ) (c o s e c ϕ + c o t ϕ) + 1 = \frac{1 - s i n ϕ + c o s ϕ}{s i n ϕ c o s ϕ} = \frac{s i n^{2} ϕ + c o s^{2} ϕ - s i n ϕ + c o s ϕ}{s i n ϕ c o s ϕ} = (c o s e c ϕ + c o t ϕ) - (s e c ϕ - t a n ϕ) = y - x$