(sec theta +tan theta)²=1+sin theta ÷1- sin theta
Answers
Answer:
Step-by-step explanation:
L.H.S. = ( sec θ + tan θ )²
= sec² θ + tan² θ + 2 sec θ.tan θ
= ( 1/cos² θ) + ( sin² θ/ cos² θ ) + 2 ( 1/cos θ) × ( sin θ / cos θ)
= ( 1/cos² θ ) + ( sin² θ/ cos² θ ) + ( 2 sin θ / cos² θ )
= ( 1 + sin θ + 2 sin θ ) / cos² θ
= ( 1 + sin θ )² / cos² θ
R.H.S. = ( 1 + sin θ ) / ( 1 - sin θ )
= [ ( 1 + sin θ ) / ( 1 - sin θ ) ] × [ ( 1 + sin θ ) / ( 1 + sin θ ) ] (∵Rationalising the denominator )
= [ ( 1 + sin θ ) × ( 1 + sin θ ) ] / [ ( 1 - sin θ ) × ( 1 + sin θ ) ]
= ( 1 + sin θ )² / ( 1 - sin²θ ) [∵(a +b)( a - b )= a² - b² ; (a + b) ( a + b ) = ( a + b )² ]
= ( 1 + sin )² θ / cos² θ [∵ 1 - sin²θ = cos² θ ]
∴ L.H.S . = R.H.S.
Hence it is proved.
Step-by-step explanation:
• 3:32 PM |
Step-by-step explanation:
L.H.S. = (sec 0 + tan 0 )²
= sec² 0 + tan² 0 + 2 sec 0.tan 0 cos ) × ( sin 0 / cos 0)
= (1/cos²0) + (sin² 0/ cos² ) + 2 ( 1/
= (1/cos² 0 ) + ( sin² 0/ cos² 0 ) + ( 2 sin
0 / cos² 0 )
= (1+ sin 0 + 2 sin 0 ) / cos² 0 = (1+ sin 0 )²/cos² 0
R.H.S. = (1+ sin 0 )/(1-sin 0) = [(1 + sin 0)/(1-sin 0 ) ] ×[(1 + sin )/(1+ sin 0 ) ] (Rationalising the denominator)
sin ) x (1 + sin 0 )] / [(1 - sin = [(1 + s 0 ) × ( 1 + sin 0 )]
= (1 + sin 0)²/(1-sin²0 ) [(a+b)( a
b)= a² - b²; (a + b)(a + b ) = (a + b )²] = (1 + sin )² 0 / cos² 0 [1-sin²0 = cos² 0]