Math, asked by bee6, 1 year ago

please tell how to prove the above statement.

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Answered by manitkapoor2
2
[tex](sin(x) + cosec(x))^{2} = sin^{2}(x) + cosec(x)^{2} + 2sin(x)cosec(x) \\ = sin^{2}(x) + cosec(x)^{2} + 2[/tex]
[tex](cos(x) + sec(x))^{2} = cos^{2}(x) + sec(x)^{2} + 2cos(x)sec(x) \\ = cos^{2}(x) + sec(x)^{2} + 2[/tex]
[tex](sin(x) + cosec(x))^{2} + (cos(x) + sec(x))^{2} = \\sin^{2}(x) + cosec(x)^{2} + 2 +\\ cos^{2}(x) + sec(x)^{2} + 2 \\ = cosec(x)^{2} + sec(x)^{2} + 1 + 4 = cosec(x)^{2} + sec(x)^{2} +5 [/tex]
[tex]cosec(x)^{2} + sec(x)^{2} + 5 = \frac{1}{sin^{2}(x)cos^{2}(x)} + 5 \\ = \frac{4}{4sin^{2}(x)cos^{2}(x)} + 5 \\ = \frac{4}{sin^{2}(2x)} + 5[/tex]

as we know that

[tex]0 \leq sin^{2}(2x) \leq 1 \\ 1 \leq \frac{1}{sin^{2}(2x)} \\ 4 + 5 \leq \frac{4}{sin^{2}(2x)} + 5\\ 9 \leq \frac{4}{sin^{2}(2x)} + 5\\ 9 \leq (sin(x) + cosec(x))^{2} + (cos(x) + sec(x))^{2}[/tex]

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