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Sagot :
Answer:
See below
Step-by-step explanation:
Here we need to prove that ,
[tex]\sf\longrightarrow sin^2\theta + cos^2\theta = 1 [/tex]
Imagine a right angled triangle with one of its acute angle as [tex]\theta[/tex] .
- The side opposite to this angle will be perpendicular .
- Also we know that ,
[tex]\sf\longrightarrow sin\theta =\dfrac{p}{h} \\[/tex]
[tex]\sf\longrightarrow cos\theta =\dfrac{b}{h} [/tex]
And by Pythagoras theorem ,
[tex]\sf\longrightarrow h^2 = p^2+b^2 \dots (i) [/tex]
Where the symbols have their usual meaning.
Now , taking LHS ,
[tex]\sf\longrightarrow sin^2\theta +cos^2\theta [/tex]
- Substituting the respective values,
[tex]\sf\longrightarrow \bigg(\dfrac{p}{h}\bigg)^2+\bigg(\dfrac{b}{h}\bigg)^2\\[/tex]
[tex]\sf\longrightarrow \dfrac{p^2}{h^2}+\dfrac{b^2}{h^2}\\ [/tex]
[tex]\sf\longrightarrow \dfrac{p^2+b^2}{h^2} [/tex]
- From equation (i) ,
[tex]\sf\longrightarrow\cancel{ \dfrac{h^2}{h^2}}\\ [/tex]
[tex]\sf\longrightarrow \bf 1 = RHS [/tex]
Since LHS = RHS ,
Hence Proved !
I hope this helps.
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