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2 years ago

5

GUYS HELP THIS IS SO IMP P;LEASE

1 answer:

ra1l [238]2 years ago

3 0

Answer:

(1+2) (2,-1)

Step-by-step explanation:

i only did number 1... i couldn't figure out the rest

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Answer :

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Answer:

See below

Step-by-step explanation:

Here we need to prove that ,

$\sf\longrightarrow sin^2\theta + cos^2\theta = 1$

Imagine a right angled triangle with one of its acute angle as $\theta$ .

The side opposite to this angle will be perpendicular .
Also we know that ,

$\sf\longrightarrow sin\theta =\dfrac{p}{h} \\$

$\sf\longrightarrow cos\theta =\dfrac{b}{h}$

And by Pythagoras theorem ,

$\sf\longrightarrow h^2 = p^2+b^2 \dots (i)$

Where the symbols have their usual meaning.

Now , taking LHS ,

$\sf\longrightarrow sin^2\theta +cos^2\theta$

Substituting the respective values,

$\sf\longrightarrow \bigg(\dfrac{p}{h}\bigg)^2+\bigg(\dfrac{b}{h}\bigg)^2\\$

$\sf\longrightarrow \dfrac{p^2}{h^2}+\dfrac{b^2}{h^2}\\$

$\sf\longrightarrow \dfrac{p^2+b^2}{h^2}$

From equation (i) ,

$\sf\longrightarrow\cancel{ \dfrac{h^2}{h^2}}\\$

$\sf\longrightarrow \bf 1 = RHS$

Since LHS = RHS ,

Hence Proved !

I hope this helps.

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