Question

If sec theta + tan theta = p, then find the value of cosec theta.

Answer 1

             

$\huge\boxed{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \huge\boxed{\bold{\csc\theta=\frac{p^2+1}{p^2-1}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ }$

$\sf{I'm taking \ \theta} \ \ \sf{as \ A. }$

$\sec^2 A -\tan^2 A=1 \\ \\ (\sec A+\tan A)(\sec A-\tan A)=1 \\ \\ p(\sec A-\tan A)=1 \\ \\ \\ \sec A-\tan A=\frac{1}{p}$

$(\sec A + \tan A)+(\sec A-\tan A)=p+\frac{1}{p} \\ \\ 2\sec A=\frac{p^2+1}{p} \\ \\ \sec A=\frac{p^2+1}{2p} \\ \\ \\ (\sec A + \tan A)-(\sec A-\tan A)=p-\frac{1}{p} \\ \\ 2\tan A=\frac{p^2-1}{p} \\ \\ \tan A=\frac{p^2-1}{2p}$

$\tan A=\frac{\sin A}{\cos A} \\ \\ \tan A=\sin A\div \cos A\\ \\ \tan A=\frac{1}{\csc A} \div \frac{1}{\sec A} \\ \\ \tan A=\frac{1}{\csc A} \times \sec A \\ \\ \tan A=\frac{\sec A}{\csc A}$

$\csc A=\frac{\sec A}{\tan A} \\ \\ \\ \csc A=\frac{\frac{p^2+1}{2p}}{\frac{p^2-1}{2p}} \\ \\ \\ \csc A=\frac{p^2+1}{2p} \div \frac{p^2-1}{2p} \\ \\ \\ \csc A=\frac{p^2+1}{2p} \times \frac{2p}{p^2-1} \\ \\ \\ \csc A=\frac{p^2+1}{p^2-1}$

$\sf{Hope this helps. \\ \\ Plz mark it as the brainliest. \\ \\ \\ Thank you. :-)}$

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Answer 2

\huge\boxed{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \huge\boxed{\bold{\csc\theta=\frac{p^2+1}{p^2-1}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ }

cscθ=

p

2

−1

p

2

+1

\sf{I'm taking \ \theta}$\ \ \sf{as \ A. }

\begin{lgathered}\sec^2 A -\tan^2 A=1 \\ \\ (\sec A+\tan A)(\sec A-\tan A)=1 \\ \\ p(\sec A-\tan A)=1 \\ \\ \\ \sec A-\tan A=\frac{1}{p} \\ \\ \\\end{lgathered}

sec

2

A−tan

2

A=1

(secA+tanA)(secA−tanA)=1

p(secA−tanA)=1

secA−tanA=

p

1

\begin{lgathered}(\sec A + \tan A)+(\sec A-\tan A)=p+\frac{1}{p} \\ \\ 2\sec A=\frac{p^2+1}{p} \\ \\ \sec A=\frac{p^2+1}{2p} \\ \\ \\ (\sec A + \tan A)-(\sec A-\tan A)=p-\frac{1}{p} \\ \\ 2\tan A=\frac{p^2-1}{p} \\ \\ \tan A=\frac{p^2-1}{2p}\end{lgathered}

(secA+tanA)+(secA−tanA)=p+

p

1

2secA=

p

p

2

+1

secA=

2p

p

2

+1

(secA+tanA)−(secA−tanA)=p−

p

1

2tanA=

p

p

2

−1

tanA=

2p

p

2

−1

\begin{lgathered}\tan A=\frac{\sin A}{\cos A} \\ \\ \tan A=\sin A\div \cos A\\ \\ \tan A=\frac{1}{\csc A} \div \frac{1}{\sec A} \\ \\ \tan A=\frac{1}{\csc A} \times \sec A \\ \\ \tan A=\frac{\sec A}{\csc A} \\ \\ \\\end{lgathered}

tanA=

cosA

sinA

tanA=sinA÷cosA

tanA=

cscA

1

÷

secA

1

tanA=

cscA

1

×secA

tanA=

cscA

secA

\begin{lgathered}\csc A=\frac{\sec A}{\tan A} \\ \\ \\ \csc A=\frac{\frac{p^2+1}{2p}}{\frac{p^2-1}{2p}} \\ \\ \\ \csc A=\frac{p^2+1}{2p} \div \frac{p^2-1}{2p} \\ \\ \\ \csc A=\frac{p^2+1}{2p} \times \frac{2p}{p^2-1} \\ \\ \\ \csc A=\frac{p^2+1}{p^2-1}\end{lgathered}

cscA=

tanA

secA

cscA=

2p

p

2

−1

2p

p

2

+1

cscA=

2p

p

2

+1

÷

2p

p

2

−1

cscA=

2p

p

2

+1

×

p

2

−1

2p

cscA=

p

2

−1

p

2

+1

\sf{Hope this helps. \\ \\ Plz mark it as the brainliest. \\ \\ \\ Thank you. :-)}\end{lgathered}

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