Question

if sec theta + tan theta + p by q then sin theta =

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:-

Given:

Sec A + tan A = p/q -- equation (1)

We know that,

Sec² ∅ - tan² ∅ = 1

→ (Sec A + tan A)(Sec A - tan A) = 1

[Since, a² - b² = (a + b)(a - b)].

Putting sec A + tan A as p/q we get,

→ (Sec A - tan A)(p/q) = 1

→ Sec A - tan A = q/p -- equation (2)

Add equations (1) & (2) we get,

→ Sec A + tan A + Sec A - tan A = p/q + q/p

→ 2Sec A = (p + q)/pq

→ Sec A = (p + q)/2pq -- equation (3)

Substitute the value of "Sec A" in equation (1)

→ Sec A + tan A = p/q

→ tan A = p/q - (p + q)/2pq

→ tan A = (2p² - p - q)/2pq -- equation (4)

Dividing equation (4) by (3) we get,

$\sf \implies \: \frac{ \tan \: a}{ \sec \: a} = \frac{ \frac{2 {p}^{2} - p - q }{2pq} }{ \frac{p + q}{2pq} } \\ \\ \sf \implies \: \frac{ \tan \: a }{ \sec \: a } = \frac{2 {p}^{2} - p - q}{p + q}$

Putting the value of tan A as Sin A/Cos A and Sec A as 1/Cos A we get,

$\sf \implies \: \frac{ \frac{ \sin \: A }{ \cos \: A } }{ \frac{1}{ \cos \: A} } = \frac{2 {p}^{2} - p - q }{p + q} \\ \\ \sf \implies \: \frac{ \sin \: A }{ \cos \: A } \times \cos \: A = \frac{2 {p}^{2} - p - q}{p + q} \\ \\ \implies\sf \large \sin \: A \: = \frac{2 {p}^{2} - p - q}{p + q}$