From the following figure- find- x0 - tan x0 x0 - tan x0

Let the given triangle be ABC, where ∠ ABC = 90^∘ , AB = 1, BC = y, AC = 2 By Pythagoras Theorem, AC^2 = AB^2 + BC^2 2^2 = 1^2 + y^2 y^2 ...

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Standard IX

Mathematics

Trigonometric Ratios

From the foll...

Question

From the following figure. find:
$(sec x^{0} - tan x^{0}) (sec x^{0} + tan x^{0})$

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Solution

Let the given triangle be ABC, where $∠ A B C = 90^{\circ}$ , $A B = 1, B C = y, A C = 2$
By Pythagoras Theorem,
$A C^{2} = A B^{2} + B C^{2}$
$2^{2} = 1^{2} + y^{2}$
$y^{2} = 3$
$y = \sqrt{3}$
Now, $(sec x^{0} - tan x^{0}) (sec x^{0} + tan x^{0})$
= ${sec}^{2} x^{\circ} - {tan}^{2} x^{\circ}$
= $(\frac{H}{B})^{2} - (\frac{P}{B})^{2}$
= $\frac{2}{1} - \frac{\sqrt{3}}{1}$
= $4 - 3$
= $1$

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In the given figure find the value of $x^{0}$

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From the following figure. find:(secx0−tanx0)(secx0+tanx0)

Let the given triangle be ABC, where ∠ABC=90∘, AB=1,BC=y,AC=2By Pythagoras Theorem,AC2=AB2+BC222=12+y2y2=3y=√3Now, (secx0−tanx0)(secx0+tanx0)= sec2x∘−tan2x∘= (HB)2−(PB)2= 21−√31= 4−3= 1

From the following figure. find:
$(sec x^{0} - tan x^{0}) (sec x^{0} + tan x^{0})$