If the value of Sin A - 1-2- then find out the value of tan A - cot A-

Given, Sin A = 1/√(2) Sin A = opposite/hypotenuse Drawing a Δ ABC, nbsp; Using pythagoras theorem, AC^2 = AB^2 + BC^2 (√(2)k)^2 = AB^2 + (1k)^2 2k^2 = AB^ ...

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Standard X

Mathematics

Trigonometric Ratios

If the value ...

Question

If the value of $S i n A = \frac{1}{\sqrt{2}}$ , then find out the value of $t a n A - c o t A$ .
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Solution

The correct option is A 0
Given,
$S i n A = \frac{1}{\sqrt{2}}$
$S i n A = \frac{o p p o s i t e}{h y p o t e n u s e}$

Drawing a $Δ A B C$ ,

Using pythagoras theorem,

$A C^{2} = A B^{2} + B C^{2}$
$(\sqrt{2} k)^{2} = A B^{2} + (1 k)^{2}$
$2 k^{2} = A B^{2} + 1 k^{2}$
$2 k^{2} - 1 k^{2} = A B^{2}$
$1 k = A B$
$A B = 1 k$

Now,
$C o s A = \frac{a d j a c e n t}{h y p o t e n u s e} = \frac{1}{\sqrt{2}}$

$t a n A = \frac{S i n A}{C o s A} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1$
$c o t A = \frac{C o s A}{S i n A} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1$

$∴ t a n A - c o t A = 1 - 1 = 0$

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If the value of Sin A=1√2, then find out the value of tan A−cot A.0

If the value of $S i n A = \frac{1}{\sqrt{2}}$ , then find out the value of $t a n A - c o t A$ .
0