Question

prove that: (cosecp-sinp) (secp-cosp) (tanp+cotp) =1

Answer 1

$\boxed{\underline{\textsf{Formulas :}}}$

$\textsf{i) sinx * cosecx = 1}$

$\textsf{ii) cosx * secx = 1}$

$\textsf{iii) tanx * cotx = 1}$

$\mathsf{iv) sin^{2}x + cos^{2}x = 1}$

$\mathsf{v) sec^{2}x - tan^{2}x = 1}$

$\mathsf{vi) cosec^{2}x - cot^{2}x = 1}$

$\boxed{\underline{\textsf{Proof :}}}$

$\textsf{Now,}$

$\small{\mathsf{(cosecp - sinp) (secp - cosp) (tanp + cotp)}}$

$=\small{\mathsf{(\frac{1}{sinp}-sinp) (\frac{1}{cosp}-cosp)(\frac{1}{cotp}+cotp)}}$

$=\mathsf{\frac{1 - sin^{2}p}{sinp} * \frac{1 - cos^{2}p}{cosp} * \frac{1 + cot^{2}p}{cotp}}$

$=\mathsf{\frac{cos^{2}p}{sinp} * \frac{sin^{2}p}{cosp} * \frac{cosec^{2}p}{cotp}}$

$=\mathsf{\frac{cos^{2}p * sin^{2}p * cosec^{2}p}{sinp * cosp * cotp}}$

$=\mathsf{\frac{cos^{2}p}{sinp * cosp * \frac{cosp}{sinp}}}$

$=\mathsf{\frac{cos^{2}p * sinp}{cos^{2}p * sinp}}$

$=\mathsf{1}$

$\to \boxed{\boxed{\tiny{\mathsf{(cosecp - sinp) (secp - cosp) (tanp + cotp) = 1}}}}$

$\textsf{Hence, proved.}$

Answer 2

Step-by-step explanation:

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