# Trigonometric ratios of 90 degree plus theta(90° + θ)

In this section you will learn trigonometric ratios of 90 degree plus theta (90° + θ) for all trigonometric ratios.
Let a revolving ray OA starting from its initial position OX, make ans angle $\angle$ XOA = $\Theta$ in a anti-clockwise direction and again the same rotating line rotates and make an angle $\angle$ AOB = $90^{0}$ in the same direction. $\angle$ XOB =$90^{0} + \Theta$. Let c and E are two points on the ray OA and OB respectively such that OC = OE = r. Draw a perpendicular from CD and EF on the X-axis.
Clearly, $\Delta$ODC and $\Delta$OFE are congruent.
OD = OF = y and CD= EF = x
So the coordinates of E are (-y,x).
 Case 1: When $\angle$ AOB is in Ist and IInd quadrant. Case 2: When $\angle$ AOB is in IInd and IIIrd quadrant. Case 3: When $\angle$ AOB is in IIIrd and IVth quadrant. Case 4: When $\angle$ AOB is in IVth and Ist quadrant.
By the definition of trigonometric ratios and $\Delta$ ODC $\cong \Delta$OFE we get,
sin ($90^{0} + \Theta ) = \frac{FE}{OE} = \frac{x}{r}$

cos $\Theta = \frac{OD}{OC} = \frac{x}{r}$ ( FE=OD and OE = OC)

∴ $\underline{sin(90^{0} + \Theta)= cos\Theta}$

cos ($90^{0} + \Theta ) = \frac{OF}{OE} = \frac{-y}{r}$

sin $\Theta = \frac{DC}{OC} = \frac{y}{r}$ ( OF=-DC and OE = OC)

∴ $\underline{cos(90^{0} + \Theta)= -sin\Theta}$

tan ($90^{0} + \Theta ) = \frac{FE}{OF} = \frac{-x}{y}$

cot $\Theta = \frac{DC}{OD} = \frac{y}{x}$ ( FE=OD and OF = -DC)

∴ $\underline{tan(90^{0} + \Theta)= cot\Theta}$

Now we know that, Csc$\Theta = \frac{1}{sin \Theta}$

∴ Csc$(90 + \Theta ) = \frac{1}{sin (90 + \Theta)}$

But $sin (90 + \Theta) = cos \Theta$

∴Csc$(90 + \Theta ) = \frac{1}{cos \Theta}$

∴ $\underline{csc(90^{0} + \Theta)= sec\Theta}$

As we know that
sec$\Theta = \frac{1}{cos \Theta}$

∴ $sec (90 + \Theta ) = \frac{1}{cos (90 + \Theta)}$

But $cos (90 + \Theta) = -sin \Theta$

∴sec $(90 + \Theta ) = -\frac{1}{sin \Theta}$

∴ $\underline{sec(90^{0} + \Theta)= -csc\Theta}$

Again,
$cot \Theta = \frac{1}{tan \Theta}$

∴ cot $(90 + \Theta ) = \frac{1}{tan (90 + \Theta}$

But $tan (90 + \Theta = -cot \Theta$

∴cot $(90 + \Theta ) = -\frac{1}{cot\Theta}$

∴ $\underline{cot(90^{0} + \Theta)= -tan\Theta}$

## Examples on trigonometric ratios of 90 degree plus theta(90° + θ)

1) Find x from the following equations :
(i) csc $(90 + \Theta) + x cos\Theta.cot(90 + \Theta) = sin(90 + \Theta)$
Solution :
csc $(90 + \Theta) + x cos\Theta.cot(90 + \Theta) = sin(90 + \Theta)$
$sec \Theta + x cos\Theta.(-tan\Theta) = cos \Theta$
$sec\Theta - x cos\Theta.\frac{sin \Theta}{cos\Theta} = cos \Theta$
$sec \Theta - xsin\Theta = cos\Theta$
$- xsin\Theta = cos\Theta - sec \Theta$
$- xsin\Theta = cos\Theta - \frac{1}{cos \Theta}$

$- xsin\Theta = \frac{cos^{2}\Theta -1}{cos \Theta}$

$- xsin\Theta = \frac{-sin^{2}\Theta}{cos \Theta}$
∴ $x = tan\Theta$

(ii) $x cot(90 + \Theta) + tan(90 + \Theta).sin\Theta + csc(90+\Theta)$= 0
Solution :
$x cot(90 + \Theta) + tan(90 + \Theta).sin\Theta + csc(90+\Theta)$= 0
$- x tan\Theta -cot\Theta.sin\Theta + sec\Theta$ = 0

$- x tan\Theta -\frac{cos\Theta}{sin\Theta}.sin\Theta + sec\Theta$ = 0

$- x tan\Theta - cos\Theta + sec\Theta$ = 0
$- x tan\Theta =cos\Theta - sec\Theta$
$- x tan\Theta =cos\Theta - \frac{1}{cos\Theta}$
$- x tan\Theta =( \frac{cos^{2}\Theta -1 }{cos\Theta})$
$- x tan\Theta =( \frac{-sin^{2}\Theta}{cos\Theta})$

$- x tan\Theta =( \frac{-sin^{2}\Theta}{cos\Theta})$

$-x = \frac{-sin^{2}\Theta}{cos\Theta}\div tan\Theta$

$- x = \frac{-sin^{2}\Theta}{cos\Theta}\div \frac{sin\Theta}{cos\Theta}$

$- x = \frac{-sin^{2}\Theta}{cos\Theta}\times \frac{cos\Theta}{sin\Theta}$

∴ $x = sin\Theta$