if cosA=2/3 find the value of 2sec^2A + 2tan^2A-9
Answers
Answer:
pythagoras theorem,
3
2
=BC
2
+2
2
BC
2
=9−4
So, BC=
5
sec
2
θ=(
2
3
)
2
=
4
9
2sec
2
θ=2×
4
9
=
2
9
tanθ=(
2
5
)
tan
2
θ=(
2
5
)
2
=
4
5
2sec
2
θ+2tan
2
θ−7
=
2
9
+2×
4
5
−7=
2
9
+
2
5
−
1
7
=
2
9+5−14
=
2
14−14
=
2
0
=0
Answer:
Value of 2sec²A + 2tan²A – 9 is (–2).
Step-by-step explanation:
cosA = 2/3
We know that,
cos = Adjacent side/Hypotenuse
So 2 units is the Adjacent side to angle A and 3 units is the Hypotenuse.
By Pythagoras Theorem,
(Hypotenuse)² = (Base)² + (Height)²
(3)² = (2)² + Height²
9 = 4 + Height²
Height² = 9 – 4
Height = √5
___________________
★ Value of 2sec²A + 2tan²A – 9
sec = Hypotenuse/Adjecent Side
= 3/2
tan = Opposite side/Adjacent Side
= √5/2
Putting their values,
⇒ 2(3/2)² + 2(√5/2)² – 9
⇒ 2(9/4) + 2(5/4) – 9
⇒ 9/2 + 5/2 – 9
⇒ 14/2 – 9
⇒ 7 – 9
⇒ –2
Therefore, value of 2sec²A + 2tan²A – 9 is (–2).