If 4sinθ=3cosθ, Then 4(1−tan2θ)sec2θ is

4sinθ=3cosθtanθ=43Now, 4(1−tan2θ)sec2θ= 4(1−tan2θ)1+ tan2θ= 4(1−(43)2)1+(43)2= 4(16−9)16+9= 2825

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If 4 =3 , Then { ^{2} }{4(1- ^{2} )} is

Question

If $4 sin θ = 3 cos θ,$ Then $\frac{sec ^{2} θ}{4 ( 1 - tan ^{2} θ )}$ is

\frac{25}{16}

\frac{25}{28}

\frac{1}{4}

\frac{16}{25}

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Alexander Discussed

4 sin θ = 3 cos θ,

Then

\frac{sec ^{2} θ}{4 ( 1 - tan ^{2} θ )}

6 mins ago

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6 mins ago

4 sin θ = 3 cos θ

tan θ = \frac{3}{4}

Now,

\frac{sec ^{2} θ}{4 ( 1 - tan ^{2} θ )}

\frac{1 + tan ^{2} θ}{4 ( 1 - tan ^{2} θ )}

\frac{1 + ( \frac{3}{4} ) ^{2}}{4 ( 1 - ( \frac{3}{4} ) ^{2} )}

\frac{16 + 9}{4 ( 16 - 9 )}

\frac{25}{28}

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Question Text	If $4 sin θ = 3 cos θ,$ Then $\frac{sec ^{2} θ}{4 ( 1 - tan ^{2} θ )}$ is
Answer Type	Text solution:1
Upvotes	150