Trigonometric ratios of 90 minus theta(−Θ)
(90 - Θ)
In this section we will discuss the relation among all trigonometric ratios of 90 minus theta (90 - $\Theta$).Let a revolving or rotating ray starts from its initial position OX and trace an angle $\angle$XOA= $\Theta$. Let P(x,y) be any point on ray OA such that OP=radius = r. Draw a perpendicular from P on the X-axis. If another ray rotate from the initial position OX and reached to OB making an angle $90^{0}$ to coincide with OY. After that it rotates in the clockwise direction through an angle $\Theta$. OA' is the final position so angle $\angle XOB= 90 - \Theta$. Let Q be a point on OB such that OP=OQ= r. Now draw a perpendicular QN in the x-axis. Then $\Delta$ OMP and
$\Delta$ ONQ are congruent.
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Case 2 : If OM and QN both are negative and similarly PM and ON are negative. Case 3 :IF QN and OM both are negative. Similarly, PM and ON are both negative. Case 3 : IF QN and OM both are positive. Similarly, ON and PM are both negative. |
So, B has coordinates (y,x)
$sin(90 - \Theta) = \frac{QN}{OQ} = \frac{x}{r} = cos \Theta$
$cos(90 - \Theta) = \frac{ON}{OQ} = \frac{y}{r} = sin \Theta$
$tan(90 - \Theta) = \frac{QN}{ON} = \frac{x}{y} = cot\Theta$
As we know that, $csc \Theta = \frac{1}{sin \Theta}$
So, $csc (90- \Theta ) = \frac{1}{sin(90- \Theta)}$
$csc (90- \Theta ) = \frac{1}{cos \Theta}$
$csc (90- \Theta ) = sec \Theta$
Similarly,
$sec \Theta = \frac{1}{cos \Theta}$
So, $sec (90- \Theta ) = \frac{1}{cos(90- \Theta)}$
$sec (90- \Theta ) = \frac{1}{sin \Theta}$
$sec (90- \Theta ) = csc \Theta$
And
$cot \Theta = \frac{1}{tan \Theta}$
So, $cot (90- \Theta ) = \frac{1}{tan(90- \Theta)}$
$cot (90- \Theta ) = \frac{1}{cot \Theta}$
$cot (90- \Theta ) = tan \Theta$
Examples on trigonometric ratios of 90 minus theta
(90 - Θ)
In a $\Delta$ ABC , prove that,1)$cos \left (\frac{A + B}{2} \right ) = sin \frac{c}{2}$
Solution : In a $\Delta$ ABC,
We know that, A + B + C = 180
⇒ A + B = 180 - c
⇒ $\left (\frac{A + B}{2} \right ) =\left (\frac{180 - c}{2} \right )$
⇒ $\left (\frac{A + B}{2} \right ) =\left (90 - \frac{c}{2} \right )$
consider, $cos \left (\frac{A + B}{2} \right )$
⇒ $cos \left (90 -\frac{c}{2} \right )$
Since $cos \left (90 -\Theta \right ) = sin \Theta$
⇒ $sin\left (\frac{c}{2} \right )$
2) $tan \left (\frac{A + B}{2} \right ) = cot \frac{c}{2}$
Solution : In a $\Delta$ ABC,
We know that, A + B + C = 180
⇒ A + B = 180 - c
⇒ $\left (\frac{A + B}{2} \right ) =\left (\frac{180 - c}{2} \right )$
⇒ $\left (\frac{A + B}{2} \right ) =\left (90 - \frac{c}{2} \right )$
consider, $tan \left (\frac{A + B}{2} \right )$
⇒ $tan \left (90 -\frac{c}{2} \right )$
Since $tan \left (90 -\Theta \right ) = cot \Theta$
⇒ $cot \left (\frac{c}{2} \right )$
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