Find 1-cos²x/1+sec²x where the perpendicular is 8cm base 6cm.
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Find 1-cos²x/1+sec²x where the perpendicular is 8cm base 6cm.
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Certainly, let's solve the trigonometric expression:
\[ \frac{1 - \cos^2(x)}{1 + \sec^2(x)} \]
First, we can simplify it by using trigonometric identities:
1. \(\sec(x) = \frac{1}{\cos(x)}\)
2. \(\cos^2(x) + \sin^2(x) = 1\)
Now, we can rewrite the expression:
\[ \frac{1 - \cos^2(x)}{1 + \sec^2(x)} = \frac{1 - \cos^2(x)}{1 + \frac{1}{\cos^2(x)}} \]
Using the identity \(\cos^2(x) + \sin^2(x) = 1\), we can rewrite \(\cos^2(x)\) as \(1 - \sin^2(x)\):
\[ \frac{1 - (1 - \sin^2(x))}{1 + \frac{1}{\cos^2(x)}} = \frac{\sin^2(x)}{\frac{1}{\cos^2(x)}} \]
Now, let's simplify further. Remember that \(\sec(x) = \frac{1}{\cos(x)}\), so \(\frac{1}{\cos^2(x)} = \sec^2(x)\):
\[ \frac{\sin^2(x)}{\sec^2(x)} \]
Now, using the definition of \(\sec(x)\) as \(\frac{1}{\cos(x)}\), we can simplify it further:
\[ \sin^2(x) \cdot \cos^2(x) \]
So, the simplified expression is \(\sin^2(x) \cdot \cos^2(x)\).
Answer:
10 cm
Step-by-step explanation:
Correct option is B)
The sides of right angled triangle are 6,8cm
The hypotenuse is given as
a
2
+b
2
=c
2
6
2
+8
2
=c
2
36+64=c
2
c
2
=100
c=
100