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Sin^2A+ sin^2B = 1 prove that ? Please provide me solution as soon as possible

True Motivator , 7 Years ago
Grade 12th pass
anser 1 Answers
Prajwal Patankar

Last Activity: 7 Years ago

Consider a right angled Triangle ABC tight abgled at B havings sides AB, BC and AC.
Now From Pythagorus theorem, we have 
(AB)2 + (BC)2 = (AC)2
 
Now Divide throughout by (AC)2 which gives
 
(AB)2 + (BC)2 = (AC)2
(AC)2    (AC)2     (AC)2
 
(AB)2 + (BC)2 = 1           ... (i)
(AC)2    (AC)2
But sinA = (BC)/(AC) 
And cosA = (AB)/(AC)
 Hence, substituting values of sinA and cosA in eq(i) , we get 
 
(cosA)2 + (sinA)2  = 1 
 
i.e. sin2A + cos2A = 1 
 
Hope it helped 👍🏻 
 
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