Question

If sec theta+tan theta = p,then find the value of sin thete interms of p

Answer 1

Step-by-step explanation:

Let theta =@

sec@ + tan@=p------------------(1)

we know ,

sec^2@-tan^2@=1

(sec@-tan@)(sec@+tan@)=1

hence ,

sec@-tan@=1/p -----------------(2)

now equation (1)and (2)

2sec@=p+1/p=(p^2+1)/p

sec@=(p^2+1)/2p

hence,

cos@=2p/(1+p^2)

hence ,

sin@=(1-p^2)/(1+p^2)

Answer 2

Answer:

Value of sinθ is $\frac{p^{2}-1}{p^{2}+1}$

Step-by-step explanation:

If secθ + tanθ = p -----(1)

By the identity sec²θ - tan²θ = 1

(secθ - tanθ)(secθ + tanθ) = 1 [Since (a - b)(a + b) = a² - b²]

(secθ - tanθ) × p = 1

secθ - tanθ = $\frac{1}{p}$ --------(2)

By adding equation (1) and equation (2)

2secθ = $p + \frac{1}{p}$

secθ = $\frac{1}{2}(\frac{p^{2}+1}{p})$

cosθ = $\frac{2p}{p^{2}+1}$

Since sinθ = $\sqrt{1-cos^{2}\theta }$

Therefore, sinθ = $\sqrt{1-(\frac{2p}{p^{2}+1})^{2}}$

sinθ = $\sqrt{\frac{(p^{4}+1+2p^{2}-4p^{2})}{(p^{2}+1)^{2}}}$

       = $\frac{1}{p^{2}+1}\sqrt{p^{4}-2p^{2}+1}$

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

       = $\frac{p^{2}-1}{p^{2}+1}$

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