If cosec theta + cot theta = p prove that cos theta = p2-1 / p2 + 1
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Answered by
6
Cosec¢+cot¢=p
=1/sin¢+cos¢/cos¢=p
=1+cos¢/sin¢=p
From Rhs
P^2-1/p^2+1
=(1+cos¢/sin¢) ^2-1/(1+sin¢/cos¢) ^2+1
=1+co¢^2¢+2cos¢-1/cos¢(1+cos^2¢+2cos¢+1/cos¢
=cos¢
Answered by
31
Answer:
Step-by-step explanation:
Cosec A + Cot A = P
= 1/Sin A + Cos A / Sin A = P
= 1+Cos A/Sin A = p
=> SQUARING ON BOTH SIDES
= (1+COS A)²/(SINA)²= P²
= (1+ Cos A)² / (Sin A)² = P²
= (1+cos A)² = (p²)[(Sin A)²]
= (1+ cos A) ² = (p²) [(1-cos²A)]
= (1+ cos A) ² = (p²) [ (1+cosA)(1-cos A) ]
= (1+cos A)² ÷ (1+ cos A) = (p²)[(1-cos A)]
= 1+cos A = (p²)[1-cos A]
= 1+cos A ÷ 1-cos A = p²
= Here, Using (a+b/a-b=c+d/c-d). This is known as Componendo and dividendo
According to the Question statement!
1+cos A ÷ 1-cos A = p²
Then,
(1+cos A) + (1 - cos A ) ÷ (1+ Sin A - (1-sin A) = p²+1 / p²-1
= 2/2cos = p²+1/p²-1
= 1/cos = p²+1/p²-1
= Sec = p²+1 /p²-1
We know that Cos A = 1/sec A
Then,
Cos A = p²-1 /p²+1
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