For positive interger n- if f n -sin n -cos n - then f 3- f 5- f 5- f 7 isA- f 1- f 3B- I 3- f 1C- f 3- f 5f 5D- f5-f7

The correct option is A f(1)f(3)Let sinθ=s, cosθ=c f(3) f(5)f(5) f(7)=s^3(1 s^2)+c^3(1 c^2)s^5(1 s^2)+c^5(1 c^2) =s^3c^2+c^3s^2s^5c^2+c^5s^2 =s+cs^3+c^3 =f( ...

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Standard XII

Mathematics

Basic Inverse Trigonometric Functions

For positive ...

Question

For positive interger $n,$ if $f (n) = {sin}^{n} θ + {cos}^{n} θ,$ then $\frac{f (3) - f (5)}{f (5) - f (7)}$ is

\frac{f (1)}{f (3)}

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\frac{f (3)}{f (1)}

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\frac{f (3)}{f (5)}

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\frac{f (5)}{f (7)}

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Solution

The correct option is A $\frac{f (1)}{f (3)}$
Let $sin θ = s, cos θ = c$
$\frac{f (3) - f (5)}{f (5) - f (7)} = \frac{s^{3} (1 - s^{2}) + c^{3} (1 - c^{2})}{s^{5} (1 - s^{2}) + c^{5} (1 - c^{2})} = \frac{s^{3} c^{2} + c^{3} s^{2}}{s^{5} c^{2} + c^{5} s^{2}} = \frac{s + c}{s^{3} + c^{3}} = \frac{f (1)}{f (3)}$

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For positive interger n, if f(n)=sinnθ+cosnθ, then f(3)−f(5)f(5)−f(7) is

The correct option is A f(1)f(3)Let sinθ=s, cosθ=c f(3)−f(5)f(5)−f(7)=s3(1−s2)+c3(1−c2)s5(1−s2)+c5(1−c2)=s3c2+c3s2s5c2+c5s2=s+cs3+c3=f(1)f(3)

For positive interger $n,$ if $f (n) = {sin}^{n} θ + {cos}^{n} θ,$ then $\frac{f (3) - f (5)}{f (5) - f (7)}$ is