N
cot (90° - 0) tano - cosec (90° - 0) seco
sin 12° cos 15° sec 78°cos ec 75°
cos? (50° + 0) + cos² (40°-0)
t an 15° tan 37° tan 53° tan 75°
Answers
Answer:
1) - 1
2) 1
3) 1
4) 1
Step-by-step explanation:
Given--->
1) Cot(90° -θ ) tanθ - Cosec(90° - θ ) Secθ
2) Sin12° Cos15° Sec78° Cosec75°
3) Cos² ( 50° + θ ) + Cos² ( 40° - θ )
4) tan15° tan37° tan53° tan75°
To find---> Value of given expression
Solution
We know that,
a) Sin(90° - θ ) = Cosθ
b) Cos( 90° - θ ) = Sinθ
c) tan ( 90° - θ ) = Cotθ
d) Cot ( 90° - θ ) = tanθ
e) Sec ( 90° - θ ) = Cosecθ
f) Cosec ( 90° - θ ) = Secθ
Now we solve the original problem one by one.
1) Cot(90° - θ ) tanθ - Cosec(90° - θ ) Secθ
Applying above formula ,
= tanθ tanθ - Secθ Secθ
= tan²θ - Sec²θ
= - ( Sec²θ - tan²θ )
We know that Sec²A - tan²A = 1 , applying it , we , get,
= - ( 1 )
= -1
2) Sin12° Cos15° Sec78° Cosec75°
= Sin12° Cos15° Sec ( 90° - 12° ) Cosec ( 90° - 15° )
= Sin12° Cos15° Cosec12° Sec15°
= Sin12° Cos15° ( 1 / Sin12° ) ( 1 / Cos15° )
= 1
3) Cos² ( 50° + θ ) + Cos² ( 40° - θ )
= Cos² ( 50° + θ ) + [ Cos { 90° - ( 50° + θ ) } ]²
= Cos² ( 50° + θ ) + { Sin ( 50° + θ ) }²
= Cos² ( 50° + θ ) + Sin² ( 50 ° + θ )
We know that , Sin² θ + Cos²θ = 1 , applying it , we get,
= 1
4) tan15° tan37° tan53° tan75°
= tan15° tan37° tan ( 90° - 37° ) tan ( 90° - 15° )
= tan15° tan37° Cot 37° Cot 15°
= tan 15° tan37° ( 1 / tan37° ) ( 1 / tan15° )
= 1
#Answerwithquality
# BAL
Answer:
Step-by-step explanation:
Given--->
1) Cot(90° -θ ) tanθ - Cosec(90° - θ ) Secθ
2) Sin12° Cos15° Sec78° Cosec75°
3) Cos² ( 50° + θ ) + Cos² ( 40° - θ )
4) tan15° tan37° tan53° tan75°
To find---> Value of given expression
Solution
We know that,
a) Sin(90° - θ ) = Cosθ
b) Cos( 90° - θ ) = Sinθ
c) tan ( 90° - θ ) = Cotθ
d) Cot ( 90° - θ ) = tanθ
e) Sec ( 90° - θ ) = Cosecθ
f) Cosec ( 90° - θ ) = Secθ
Now we solve the original problem one by one.
1) Cot(90° - θ ) tanθ - Cosec(90° - θ ) Secθ
Applying above formula ,
= tanθ tanθ - Secθ Secθ
= tan²θ - Sec²θ
= - ( Sec²θ - tan²θ )
We know that Sec²A - tan²A = 1 , applying it , we , get,
= - ( 1 )
= -1
2) Sin12° Cos15° Sec78° Cosec75°
= Sin12° Cos15° Sec ( 90° - 12° ) Cosec ( 90° - 15° )
= Sin12° Cos15° Cosec12° Sec15°
= Sin12° Cos15° ( 1 / Sin12° ) ( 1 / Cos15° )
= 1
3) Cos² ( 50° + θ ) + Cos² ( 40° - θ )
= Cos² ( 50° + θ ) + [ Cos { 90° - ( 50° + θ ) } ]²
= Cos² ( 50° + θ ) + { Sin ( 50° + θ ) }²
= Cos² ( 50° + θ ) + Sin² ( 50 ° + θ )
We know that , Sin² θ + Cos²θ = 1 , applying it , we get,
= 1
4) tan15° tan37° tan53° tan75°
= tan15° tan37° tan ( 90° - 37° ) tan ( 90° - 15° )
= tan15° tan37° Cot 37° Cot 15°
= tan 15° tan37° ( 1 / tan37° ) ( 1 / tan15° )
= 1