. If P, Q and R are angle of triangle PQR then prove that, cos ec (P+R/2) = secQ/2
. If P, Q and R are angle of triangle PQR then prove that, cos ec (P+R/2) = secQ/2
If P, Q and R are the angles of triangle PQR, then cosec((P+R)/2) = sec(Q/2)
Since P, Q and R are the angles of triangle, then they hold the relation
P + Q + R = 180° .....(i)
Rearranging this equation, we get
P + R = 180° - Q ---(ii)
Using the lhs of the equation,
cosec((P+R)/2)
Substituting (P+R) from (ii), we get
cosec((180°-Q)/2)
=> cosec((180/2)°- (Q/2))
=> cosec(90°- (Q/2))
We know that cosec(90°- A) = sec(A). Using this in the above relation, we get
=> sec(Q/2)
which equates to the rhs of the equation given the question.
Therefore, cosec((P+R)/2) = sec(Q/2)
To learn more about trigonometry,
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