\operatorname { sec } \theta + \operatorname { tan } \theta = p , \tex

1.	\operatorname { sec } \theta + \operatorname { tan } \theta = p , \text { show that } \frac { p ^ { 2 } - 1 } { p ^ { 2 } + 1 } = \operatorname { sin } \theta
Answer» replace p → k secx+tanx=k [1/cosx+sinx/cosx]=k (1+sinx)/cosx=k (1+sinx)=kcosx (1+sinx)²=k²cos²k 1+2sinx+sin²x=k²(1-sin²x) 1+2sinx+sin²x=k²-k²sin²x (k²+1)sin²x+2sinx+1-k²=0 use the quadratic formula to solve: sinx=t (k²+1)t²+2t+1-k²=0 -2±√(2)²-4(k²+1)(1-k²) / 2k²+2 -2±√4k⁴/ 2k²+2 =>-2±2k² / 2k²+2 =>-1±k² / k²+1 => -1-k² / k²+1 or -1+k² / k²+1 In conclusion sinx=t =>sinx= (k²-1)/(k²+1)

\operatorname { sec } \theta + \operatorname { tan } \theta = p , \text { show that } \frac { p ^ { 2 } - 1 } { p ^ { 2 } + 1 } = \operatorname { sin } \theta

Answer»

replace p → k

secx+tanx=k

[1/cosx+sinx/cosx]=k

(1+sinx)/cosx=k

(1+sinx)=kcosx

(1+sinx)²=k²cos²k

1+2sinx+sin²x=k²(1-sin²x)

1+2sinx+sin²x=k²-k²sin²x

(k²+1)sin²x+2sinx+1-k²=0

use the quadratic formula to solve:

sinx=t

(k²+1)t²+2t+1-k²=0

-2±√(2)²-4(k²+1)(1-k²) / 2k²+2

-2±√4k⁴/ 2k²+2 =>-2±2k² / 2k²+2 =>-1±k² / k²+1 =>

-1-k² / k²+1 or -1+k² / k²+1

In conclusion sinx=t =>sinx= (k²-1)/(k²+1)