tan theta + sec theta-1/tan theta - sec theta+1 = 1+sin theta/cos theta
Answers
Answer:
To Prove:
tanθ−secθ+1
tanθ+secθ−1
=
cosθ
1+sinθ
Solution:
L.H.S =
tanθ−secθ+1
tanθ+secθ−1
We can write, sec
2
θ−tan
2
θ=1
=
tanθ−secθ+1
tanθ+secθ−(sec
2
θ−tan
2
θ)
=
tanθ−secθ+1
tanθ+secθ−(secθ−tanθ)(secθ+tanθ)
=
tanθ−secθ+1
(tanθ+secθ){1−(secθ−tanθ)}
=
tanθ−secθ+1
(tanθ+secθ){1−secθ+tanθ}
=tanθ+secθ
=
cosθ
sinθ
+
cosθ
1
=
cosθ
1+sinθ
= R.H.S
since L.H.S = R.H.S
tanθ−secθ+1
tanθ+secθ−1
=
cosθ
1+sinθ
Hence Proved.
Answer:
Step-by-step explanation:
L.H.S. = ( tan θ + sec θ - 1 ) / ( tan θ - sec θ + 1 )
= ( tan θ - 1 + sec θ ) / ( tanθ + 1 - sec θ )
=[ tan θ - ( 1 - sec θ ) ] / [tan θ + ( 1 - sec θ ) ]
= [ tan θ - ( 1 - sec θ ) ] × [ (tan θ - ( 1 - sec θ )] / [tan θ + ( 1 - sec θ )] × [tan θ - ( 1- sec θ ) ] ( ∵ Rationalising the denominatdenominator)
= [ tan θ - ( 1 - sec θ ) ]² / [ tan² θ - ( 1 - sec θ )² ] [∵( a-b)² = (a-b)(a-b) ; a²-b² = (a-b) (a+b) ]
= [ tan²θ + ( 1 - sec θ)² - 2 ( tan θ ) ( 1 - sec θ ) ] / [tan² θ - ( 1 + sec² θ + 2sec θ ) ]
[ ∵ ( a +b )² = a² + b² + 2ab ]
= [ tan² θ + 1 + sec² θ - 2 sec θ - 2 tanθ + 2 tan θ sec θ ] / [ tan² θ - 1 - sec² θ + 2 sec θ ]
= [ sec² θ + sec² θ - 2 sec θ - 2 tan θ + 2 tan θ sec θ ] / [ tan² θ - sec² θ - 1 + 2 sec θ ] [∵ sec²θ = tan² θ + 1 ]
= [ 2 sec² θ + 2tan θ sec θ - 2 sec θ - 2 tan θ ] / [ - 1 - 1 + 2 sec θ ]
[ ∵tan² θ - sec² θ = - 1 ]
= [ 2 sec θ ( sec θ + tan θ ) - 2 ( sec θ + tan θ ) ] / [ - 2 + 2 sec θ ]
= [ ( sec θ + tan θ ) ( 2 sec θ - 2 ) ] / [ 2 sec θ - 2 ]
= sec θ + tan θ
= ( 1 / cos θ ) + ( sin θ / cos θ ) [∵sec θ = 1 / cos θ ; tan θ = sin θ / cos θ ]
= ( 1 + sin θ ) / cos θ ( ∵ lcm is cos θ )
= R.H.S.
L.H.S. = R.H.S.
Hence, it is proved.