Question

if cosecant theta + cot theta is equals to p show that p square + 1 by p square minus 1 is equal to secant theta​

Answer 1

Answer:

$\displaystyle\bf\frac{p^2+1}{p^2-1}=sec\theta$

Step-by-step explanation:

Given:

$p=cosec\theta+cot\theta$.......(1)

$\frac{1}{p}=\frac{1}{cosec\theta+cot\theta}$

$\displaystye\frac{1}{p}=\frac{1}{cosec\theta+cot\theta}\times\,\frac{cosec\theta-cot\theta}{cosec\theta-cot\theta}$

$\displaystyle\frac{1}{p}=\frac{cosec\theta-cot\theta}{cosec^2\theta-cot^2\theta}$

Using

$\boxed{cosec^2A-cot^2A=1}$

$\displaystyle\frac{1}{p}=\frac{cosec\theta-cot\theta}{1}$

$\displaystyle\frac{1}{p}=cosec\theta-cot\theta$..........(2)

Adding (1) and (2)

$\displaystyle\,p+\frac{1}{p}=2\,cosec\theta$......(3)

subtracting (2) from (1)

$\displaystyle\,p-\frac{1}{p}=2\,cot\theta$.......(4)

(3) divided by (4)

$\displaystyle\frac{p+\frac{1}{p}}{p-\frac{1}{p}}=\frac{2\,cosec\theta}{2cot\theta}$

$\displaystyle\frac{\frac{p^2+1}{p}}{\frac{p^2-1}{p}}=\frac{cosec\theta}{cot\theta}$

$\displaystyle\frac{p^2+1}{p^2-1}=\frac{1/sin\theta}{cos\theta/sin\theta}$

$\displaystyle\frac{p^2+1}{p^2-1}=\frac{1}{cos\theta}$

$\displaystyle\implies\boxed{\bf\frac{p^2+1}{p^2-1}=sec\theta}$

Answer 2

Answer with Step-by-step explanation:

Given :$Cosec\theta+cot\theta=p$

To prove that $\frac{p^2+1}{p^2-1}=sec\theta$

Proof:

$cosec\theta+cot\theta=\frac{1}{sin\theta}+\frac{cos\theta}{sin\theta}$

$cosec\theta=\frac{1}{sin\theta},cot\theta=\frac{cos\theta}{sin\theta}$

$=\frac{1+cos\theta}{sin\theta}$

$\frac{p^2+1}{p^2-1}=\frac{(\frac{1+cos\theta}{sin\theta})^2+1}{(\frac{1+cos\theta}{sin\theta})^2-1}$

$\frac{p^2+1}{p^2-1}=\frac{\frac{(1+cos\theta)^2}{sin^2\theta}+1}{\frac{(1+cos\theta)^2}{sin^2\theta}-1}$

$\frac{p^2+1}{p^2-1}=\frac{\frac{(1+cos\theta)^2}{1-cos^2\theta}+1}{\frac{(1+cos\theta)^2}{1-cos^2\theta}-1}$

Using formula:$sin^2\theta=1-cos^2\theta$

$\frac{p^2+1}{p^2-1}=\frac{\frac{(1+cos\theta)^2}{(1-cos\theta)(1+cos\theta)}+1}{\frac{(1+cos\theta)^2}{(1+cos\theta)(1-cos\theta)}-1}$

Using identity: $a^2-b^2=(a+b)(a-b)$

$\frac{p^2+1}{p^2-1}=\frac{\frac{1+cos\theta}{1-cos\theta}+1}{\frac{1+cos\theta}{1-cos\theta}-1}$

$\frac{p^2+1}{p^2-1}=\frac{\frac{1+cos\theta+1-cos\theta}{1-cos\theta}}{\frac{1+cos\theta-1+co\theta}{1-cos\theta}}$

$\frac{p^2+1}{p^2-1}=\frac{2}{2cos\theta}=\frac{1}{cos\theta}$

$\frac{p^2+1}{p^2-1}=sec\theta$

Using formula:$sec\theta=\frac{1}{cos\theta}$

Hence, proved.

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