Question

Sectheta + tantheta = p , show that sectheta - tantheta = 1/p. Hence , find the values of costheta and sintheta

Answer 1

$\textbf{Given:}$

$\bf\,p=sec\,\theta+tan\,\theta$........(1)

$\frac{1}{p}=\frac{1}{sec\,\theta+tan\,\theta}{\times}\frac{sec\,\theta-tan\,\theta}{sec\,\theta-tan\,\theta}$

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

$\text{Using, \bf\,sec^A-tan^2A=1 }$

$\frac{1}{p}=\frac{sec\,\theta-tan\,\theta}{1}$

$\bf\frac{1}{p}=sec\,\theta-tan\,\theta$.................(2)

$\text{Adding (1) and (2)}$

$p+\frac{1}{p}=sec\,\theta+tan\,\theta+sec\,\theta-tan\,\theta$

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

$\implies\bf\,sec\,\theta=\frac{p^2+1}{2p}$...........(3)

$\text{Taking recirocals, we get}$

$\implies\bf\,cos\,\theta=\frac{2p}{p^2+1}$

$\text{Subtracting (2) from (1)}$

$p-\frac{1}{p}=sec\,\theta+tan\,\theta-sec\,\theta+tan\,\theta$

$\implies\frac{p^2-1}{p}=2\,tan\,\theta$

$\implies\bf\,tan\,\theta=\frac{p^2-1}{2p}$...........(4)

$\text{Dividing (4) by (3)}$

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

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

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

$\therefore\boxed{\bf\,cos\,\theta=\frac{2p}{p^2+1}\;\&\;sin\,\theta=\frac{p^2-1}{p^2+1}}$

Answer 2

SecФ - tanФ  = $\dfrac{1}{p}$  is Proved  

The value of CosФ = $\dfrac{2p}{p^{2}+1 }$  ,  SinФ  = $\dfrac{(p^{2}-1) }{(p^{2}+1 ) }$

Step-by-step explanation:

Given as :

The trigonometrical equation

SecФ + tanФ  = p              .........1

To prove  : SecФ - tanФ  = $\dfrac{1}{p}$                ........2

According to question

∵  SecФ + tanФ  = p

Or,  $\dfrac{1}{Sec\Theta +tan\Theta }$   = $\dfrac{1}{p}$

Or,  ( $\dfrac{1}{Sec\Theta +tan\Theta }$ ) ( $\dfrac{Sec\Theta -tan\Theta}{Sec\Theta -tan\Theta }$ ) = $\dfrac{1}{p}$

Or,   $\dfrac{Sec\Theta -tan\Theta}{Sec^{2} \Theta -tan^{2} \Theta }$  =  $\dfrac{1}{p}$

Or,  $\dfrac{Sec\Theta -tan\Theta}{1}$   =  $\dfrac{1}{p}$                       (∵ Sec²Ф  - tan²Ф  = 1 )

∴  SecФ - tanФ  = $\dfrac{1}{p}$   Proved

Again

Adding 1 and 2

 ( SecФ + tanФ ) + ( SecФ - tanФ ) = p + $\dfrac{1}{p}$

Or,  ( SecФ  + SecФ ) + ( tanФ - tanФ ) = p + $\dfrac{1}{p}$

Or,   2 SecФ   = $\dfrac{p^{2}+ 1}{p}$

Or,    SecФ   = $\dfrac{p^{2}+ 1}{2p}$

∵     SecФ  =  $\dfrac{1}{Cos\Theta }$

So,    $\dfrac{1}{Cos\Theta }$ = $\dfrac{p^{2}+ 1}{2p}$

By cross multiplication

 CosФ = $\dfrac{2p}{p^{2}+1 }$

Again

∵      Sin²Ф  +  Cos²Ф  = 1

So,    Sin²Ф   = 1 -  Cos²Ф

Put the value of CosФ

i.e   Sin²Ф   = 1 -  ( $\dfrac{2p}{p^{2}+1 }$ )²

Or,  Sin²Ф = $\dfrac{(p^{2}+1) ^{2} -4p^{2} }{(p^{2}+1 )^{2} }$

Or,  Sin²Ф = $\dfrac{(p^{2}-1) ^{2} }{(p^{2}+1 )^{2} }$

∴    SinФ  = $\dfrac{(p^{2}-1) }{(p^{2}+1 ) }$

Hence, SecФ - tanФ  = $\dfrac{1}{p}$  is Proved  

And The value of CosФ = $\dfrac{2p}{p^{2}+1 }$  ,  SinФ  = $\dfrac{(p^{2}-1) }{(p^{2}+1 ) }$  Answer