a=√3/2; solve √1+a + √1-a
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Answered by
2
GIVEN: a=√3/2
TO PROVE:√1+a + √1-a
Consider,
√1+a=√1+√3/2
=√1 + √√3/2
=1+3/2
=5/2
√1-a=√1-√3/2
=√1 - √√3/2
=1 - 3/2
=2-3 / 2
=-1/2
therefore,
√1+a + √1-a= 5/2 +(-1/2)
=5/2 - 1/2
=5-1 / 2
=4/2
√1+a + √1-a=2
TO PROVE:√1+a + √1-a
Consider,
√1+a=√1+√3/2
=√1 + √√3/2
=1+3/2
=5/2
√1-a=√1-√3/2
=√1 - √√3/2
=1 - 3/2
=2-3 / 2
=-1/2
therefore,
√1+a + √1-a= 5/2 +(-1/2)
=5/2 - 1/2
=5-1 / 2
=4/2
√1+a + √1-a=2
Answered by
5
the value of √1+a + √1-a is 2
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