4. If tano = 1/7. Show that: (cosec^2 0-sec^2 0)/(cosec^2 0 + sec²0) = 3/4
Answers
Step-by-step explanation:
Given :-
Tan θ = 1/7
Correction :-
Tan θ = 1/√7
To find :-
Show that :
(Cosec² θ - Sec² θ )/( Cosec² θ + Sec² θ) = 3/4
Solution :-
Method -1:-
Given that
Tan θ = 1/√7 ----------(1)
Consider a right angled triangle ABC
Right angle is at B
Now ,
Tan A = Opposite side to A/ Adjacent side to A
Now,
Tan θ= AB/BC
=> AB / BC= 1/√7
Let AB = k and BC =√ 7k
By Pythagorous theorem
AC² = AB²+BC²
=> AC² = k²+(√7k)²
=> AC² = k²+7k²
=> AC² = 8k²
=>AC = √(8k²)
=> AC = √8k
Now,
Cosec θ = Hypotenuse / Opposite side to θ
=> Cosec θ = √8 k/k
=> Cosec θ = √8
=> Cosec² θ = (√8)²
=> Cosec² θ = 8
and Sec θ = Hypotenuse / Adjacent side to θ
=> Sec θ = √8 k / √7k
=> Sec θ = √8/√7
=> Sec² θ = (√8/√7)²
=> Sec² θ = 8/7
Now ,
LHS:-
(Cosec² θ - Sec² θ )/( Cosec² θ + Sec² θ)
=> [8-(8/7)] / [ 8+(8/7)]
=> [(8×7-8)/7] / [ 8×7+8)/7]
=> [(56-8)/7] / [ (56+8)/7]
=> (48/7)/(64/7)
=> (48/7)×(7/64)
=> 48/64
=> (3×16)/(4×16)
=> 3/4
=> RHS
LHS = RHS
Hence, Proved.
Method-2:-
Given that
Tan θ = 1/√7 ----------(1)
On squaring both sides then
=> Tan² θ = (1/√7)²
=> Tan² θ = 1/7
On adding 1 both sides then
=> 1+ Tan² θ = 1+( 1/7)
=> 1+ Tan² θ = (7+1)/7
=> 1+ Tan² θ = 8/7
We know that
Sec² θ - Tan² θ = 1
=> Sec² θ = 8/7 ---------(2)
On taking (1)
Tan θ = 1/√7
=> 1/ Tan θ = √7
=> Cot θ = √7
On squaring both sides then
=> Cot² θ = (√7)²
=> Cot² θ = 7
On adding 1 both sides then
=> 1+ Cot² θ = 1+ 7
=> 1+ Cot² θ = 8
We know that
Cosec² θ - Cot² θ = 1
=> Cosec² θ = 8 -----------(3)
Now ,
LHS:-
(Cosec² θ - Sec² θ )/( Cosec² θ + Sec² θ)
From (2)&(3)
=> 8-(8/7)] / [ 8+(8/7)]
=> [(8×7-8)/7] / [ 8×7+8)/7]
=> [(56-8)/7] / [ (56+8)/7]
=> (48/7)/(64/7)
=> (48/7)×(7/64)
=> 48/64
=> (3×16)/(4×16)
=> 3/4
=> RHS
LHS = RHS
Hence, Proved.
Used formulae:-
- Tan A = Opposite side to A/ Adjacent side to A
- Cosec θ = Hypotenuse /Opposite side to θ
- Sec θ = Hypotenuse/ Adjacent side to θ
- Sec² θ - Tan² θ = 1
- Cosec² θ - Cot² θ = 1
- Cot θ = 1/Tan θ
Pythagoras theorem:-
- In a right angled triangle, The square of the hypotenuse is equal to the sum of the squares of the other two sides.
Note :-
If Tan θ = 1/7 then we get the value of
(Cosec² θ - Sec² θ )/( Cosec² θ + Sec² θ) is 24/25 not 3/4.
