Question

sec squared theta + cosec squared theta is greater than equal to 4

Answer 1

$\textbf{To prove:}$

$sec^2\theta+cosec^2\theta\;{\geq}\;4$

$\textbf{Solution:}$

$\text{We know that,}$

$\text{For all values of}\;\theta,\;-1\;{\leq}\;sin\theta\;{\leq}\;1$

$\implies\,sin^2\theta\;{\leq}\;1$

$\implies\,\dfrac{1}{sin^2\theta}\;{\geq}\;1$

$\text{Now,}$

$sec^2\theta+cosec^2\theta$

$=\dfrac{1}{cos^2\theta}+\dfrac{1}{sin^2\theta}$

$=\dfrac{sin^2\theta+cos^2\theta}{cos^2\theta\,sin^2\theta}$

$=\dfrac{1}{cos^2\theta\,sin^2\theta}$

$\text{Multiply both numerator and denominator by 4}$

$=\dfrac{4}{4\,cos^2\theta\,sin^2\theta}$

$=\dfrac{4}{4\,sin^2\theta\,cos^2\theta}$

$=\dfrac{4}{(2\,sin\theta\,cos\theta)^2}$

$=\dfrac{4}{(sin\,2\theta)^2}$

$=4{\times}\dfrac{1}{sin^22\theta}$

${\geq}\;4(1)$

${\geq}\;4$

$\textbf{Answer:}$

$\implies\boxed{\bf\,sec^2\theta+cosec^2\theta\;{\geq}\;4}$