Question

sec x + tan x = k then show that sin x = k^2 - 1 / k^2 + 1

Answer 1

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 

Answer 2

hello users .....

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we have given that :-
sec x + tan x = k

we have to show that :-
sin x = (k² -1) / (k² +1)

solution:-
we know that :
sec²x - tan²x = 1

Here,
sec x + tan x = k
=> tan x = ( k - sec x )
=> tan²x = (k - sec x )²
.................... squaring on both side............

=> tan²x = (k² +sec²x - 2k sec x )

=> sec²x - tan²x = 2k sec x - k²

=> 1 + k² = 2k sec x
......( sec²x - tan²x = 1 ) ....

=> (1 +k²) / 2k = sec x

=> cos x = 1 / sec x = 2k / (k²+1)

now,
we know that:-
sin x = √(1-cos²x)

here,
sin x = √[ 1 - { (2k)²/ (k²+1)² } ]

=> √ [ { (k² +1)² - 4k² } / (k²+1)² ]

......(taking L.C.M ).........

=> √ [ (k² +1²+2k² - 4k² ) / (k²+1)² ]

=> √ [ (k² +1² - 2k²) / (k²+1)² ]

=> √ (k²-1)² / (k² +1)²

=> (k² -1) / (k²+1)

Hence;
sin x = (k²-1) / (k²+1)

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⭐✡ hope it helps ✡⭐