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If sinθ+cosecθ=4 Then sin^2θ-cosec^2θ =?

1 Answer

maganbhai P. · 1s2s2p

Apr 27, 2018

${sin}^{2} θ - {csc}^{2} θ = - 8 \sqrt{3}$ $sin^2theta-csc^2theta=-8sqrt3$

Explanation:

Here,

If $sinθ+cosecθ=4$ , then $sin^2θ-cosec^2θ =?$

Let

$color(blue)(sintheta+csctheta=4...to(1)$

Squaring both sides

$(sintheta+csctheta)^2=4^2$

$=>sin^2theta+2sinthetacsctheta+csc^2theta=16$

$=>sin^2theta+csc^2theta=16-2sinthetacsctheta$

Adding , $color(green)(-2sinthetacsctheta$ both sides

$sin^2theta-2sinthetacsctheta+csc^2theta=16- 4sinthetacsctheta$

$(sintheta-csctheta)^2=16-4 ,where, color(green)(sinthetacsctheta=1$

$(sintheta-csctheta)^2=12=(4xx3)=(2sqrt3)^2$

$sintheta-csctheta=+-2sqrt3$

But, $color(red)(-1 <= sintheta <= 1 and sintheta+csctheta=4$

$:.color(red)(1 <= csctheta <= 4=>sintheta < csctheta=>sintheta-csctheta < 0$

So,

$color(blue)(sintheta-csctheta=-2sqrt3...to(2)$

From $color(blue)((1)and(2)$ ,we get

$sin^2theta-csc^2theta=(sintheta+csctheta)(sintheta-csctheta)$

$sin^2theta-csc^2theta=(4)(-2sqrt3)$

$sin^2theta-csc^2theta=-8sqrt3$

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