F
Draw a circle of radius 3 cm. Take a point P outside the circle at a
distance of 7 cm from the centre o of the circle and draw two tangents to
the circle.
Answers
Answer:
Answer:
\LARGE{\bf{\underline{\underline{GIVEN:-}}}}
GIVEN:−
\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2} < /p > < p >∙
(1+sinA+cosA)
2
(1+sinA−cosA)
2
</p><p>
\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}
SOLUTION:−
LHS:
\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}→
(1+sinA+cosA)
2
(1+sinA−cosA)
2
Expand the fractions using .
\sf \to \dfrac{(cos^2-2sincos+sin^2-2cos+2sin+1)}{(cos^2+2sincos+sin^2+2cos+2sin+1)}→
(cos
2
+2sincos+sin
2
+2cos+2sin+1)
(cos
2
−2sincos+sin
2
−2cos+2sin+1)
Rearrange the terms.
\sf \to \dfrac{(cos^2+sin^2-2sincos-2cos+2sin+1)}{(cos^2+sin^2+2sincos+2cos+2sin+1)}→
(cos
2
+sin
2
+2sincos+2cos+2sin+1)
(cos
2
+sin
2
−2sincos−2cos+2sin+1)
We know that cos²A+sin²A=1.
\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→
2sin+1
1−2sincos−2cos
Now here, take -2cos common from the numerator and +2cos common from the denominator.
\sf \to \dfrac{1-2cos(sin+2)}{2sin+1}→
2sin+1
1−2cos(sin+2)
Now, rearrange the terms, add 1 and 1 and take 2 common.
Answer:
Given: Radius of the circle=3cm, OP=7cm.
Construction:
(i) With O as the centre draw a circle of radius 3cm.
(ii) Mark a point P at a distance of 7cm from O and join OP.
(iii) Draw the perpendicular bisector of OP. Let it meet OP at M.
(iv) With M as centre and MO as radius, draw another circle.
(v) Let the two circles intersect at T and T'.
(vi) Join PT and PT'. They are the required tangents. Length of the tangent, PT=6.3cm.
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