Question

answer my question plzz...
plzz fast...​

Answer 1

Hello mate ,

answer your answer in attachment,

Hope it helps

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Answer 2

Given :

sec Ф + tan Ф = p

Use sec Ф = 1 / cos Ф and tan Ф as sin Ф / cos Ф :

⇒ 1/cos Ф + sin Ф / cos Ф = p

⇒ ( 1 + sin Ф ) / cos Ф = p

Square both sides :

⇒  ( 1 + sin Ф )² / cos² Ф = p²

⇒ ( 1 + sin Ф )² = p² cos²Ф

Use cos²Ф = 1 - sin²Ф :

⇒ ( 1 + sin Ф )² = p² - p² sin²Ф

⇒ 1 + sin²Ф + 2 sin Ф = p² - p² sin²Ф

⇒ sin²Ф + p²sin²Ф + 2 sinФ + 1 = 0

⇒ sin²Ф ( 1 + p² ) + 2 sin Ф + 1 - p = 0

Compare with ax² + bx + c = 0

a = p² + 1

b = 2

c = 1

Using the quadratic formula we can get the value of sin Ф :

$sin\theta=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\implies \frac{-2\pm\sqrt{4-4\times(p^2+1)(1-p^2)}}{2(p^2+1)}\\\\\implies \frac{-2\pm\sqrt{4-4(1-p^4)}}{2(p^2+1)}\\\\\implies \frac{-2\pm2\sqrt{1-1+p^4}}{2(p^2+1)}\\\\\implies \frac{-2\pm 2 p^2}{2(p^2+1)}\\\\\\\implies \frac{-1\pm p^2}{p^2+1}$

cosec Ф = 1/sin Ф

$\implies \frac{p^2+1}{-1\pm p^2}\\\\Either :\frac{p^2+1}{-1-p^2}\\\\\implies\boxed{ -1}\\\\Or:\frac{p^2+1}{-1+p^2}\\\\\implies\boxed{\frac{p^2+1}{p^2-1}}$