If tanQ +sinQ=m and tanQ-sinQ=n PT (m2-n2)=4√mn
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given PT m^2-n^2=4√mn
LHS=m^2-n^2
=(tanQ+sinQ)^2-(tanQ-sinQ)^2
=(tanQ)^2 + (sinQ)^2+ 2tanQsinQ-{(tanQ)^2 + (sinQ)^2- 2tanQsinQ}
=(tanQ)^2 + (sinQ)^2+ 2tanQsinQ-(tanQ)^2 - (sinQ)^2+ 2tanQsinQ
=4tanQsinQ
RHS=4√mn
=4√(tanQ+sinQ)(tanQ-sinQ)
=4√(tanQ)^2 -(sinQ)^2 *
=4√{(sinQ)^2/(cosQ)^2}-(sinQ)^2
=4√{(sinQ)^2-(sinQ)^2(cosQ)^2}/(cosQ)^2
=4√{(sinQ)^2-(sinQ)^2{1-(sinQ)^2}/(cosQ)^2
=4√{(sinQ)^2-(sinQ)^2+(sinQ)^4}/(cosQ)^2
=4√(sinQ)^4}/(cosQ)^2
=4(sinQ)^2/(cosQ)
=4(tanQ)(sinQ)
Since ,LHS=RHS
Hence proved.
LHS=m^2-n^2
=(tanQ+sinQ)^2-(tanQ-sinQ)^2
=(tanQ)^2 + (sinQ)^2+ 2tanQsinQ-{(tanQ)^2 + (sinQ)^2- 2tanQsinQ}
=(tanQ)^2 + (sinQ)^2+ 2tanQsinQ-(tanQ)^2 - (sinQ)^2+ 2tanQsinQ
=4tanQsinQ
RHS=4√mn
=4√(tanQ+sinQ)(tanQ-sinQ)
=4√(tanQ)^2 -(sinQ)^2 *
=4√{(sinQ)^2/(cosQ)^2}-(sinQ)^2
=4√{(sinQ)^2-(sinQ)^2(cosQ)^2}/(cosQ)^2
=4√{(sinQ)^2-(sinQ)^2{1-(sinQ)^2}/(cosQ)^2
=4√{(sinQ)^2-(sinQ)^2+(sinQ)^4}/(cosQ)^2
=4√(sinQ)^4}/(cosQ)^2
=4(sinQ)^2/(cosQ)
=4(tanQ)(sinQ)
Since ,LHS=RHS
Hence proved.
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=
- Sine and cosine are trigonometric functions of an angle in mathematics. In the context of a right triangle, the sine and cosine of an acute angle are defined as the ratio of the length of the side directly opposite the angle to the length of the longest side of the triangle (the hypotenuse), and the neighbouring leg's length to the hypotenuse, respectively.
- The definitions of sine and cosine can be expanded more broadly to include any real value in terms of the lengths of certain line segments in a unit circle. The sine and cosine can be extended to arbitrary positive and negative values as well as complex numbers according to more recent definitions that represent them as infinite series or as the solutions to certain differential equations.
Here, according to the given information, we are given that,
Also,
Now, by utilizing the algebraic identity that is,
and
Now,
Also, we get that,
Also,
Then, = 4(tanQsinQ) =
Hence, =
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