Sin teta minus cos teta plis 1 divided by sin teta plus cos teta minus 1 equals 1 divided by sec teta minus tan teta
Answers
sinΘ - cosΘ + 1
_____________ = 1 /secΘ - tanΘ
sinΘ + cosΘ - 1
Taking LHS
sinΘ - cosΘ + 1
= _____________
(sinΘ + cosΘ) - 1
sinΘ - cosΘ + 1 (sinΘ + cosΘ) + 1
= _____________ × ________________
(sinΘ + cosΘ) - 1 (sinΘ + cosΘ) + 1
(Taking Conjugate)
(sinΘ - cosΘ + 1)(sinΘ + cosΘ + 1)
= _____________________________
(sinΘ + cosΘ - 1)(sinΘ + cosΘ + 1)
sin²Θ + sinΘcosΘ + sinΘ - sinΘcosΘ - cos²Θ - cosΘ + sinΘ + cosΘ + 1
= ________________________________________________________
sin²Θ + sinΘcosΘ + sinΘ + sinΘcosΘ + cos²Θ + cosΘ - sinΘ - cosΘ - 1
Cancelling out,
sin²Θ + 2sinΘ + (1 - cos²Θ)
= __________________________
(sin²Θ + cos²Θ) - 1 + 2sinΘcosΘ
sin²Θ + 2sinΘ + sin²Θ
= ___________________ (By sin²Θ + cos²Θ =1)
1 - 1 + 2sinΘcosΘ
2sin²Θ + 2sinΘ
= _____________
2sinΘcosΘ
2sinΘ(sinΘ+1)
= _____________
2sinΘ(cosΘ)
sinΘ + 1
= ________
cosΘ
= tanΘ + secΘ ...............(1)
Taking RHS
1
= ________
tanΘ - secΘ
1 tanΘ + secΘ
= ________ × __________
tanΘ - secΘ tanΘ + secΘ
(Taking Conjugate)
tanΘ + secΘ
= _____________
tan²Θ - sec²Θ
= tanΘ + secΘ ................(2) (By sec²Θ = 1 + tan²Θ)
From (1) and (2)
We get, LHS = RHS
HENCE PROVED
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