Question

1- Sino÷
1 + Sino=
(sec o t tan o )^2

वन माइनस साइन थीटा डिवाइडेड बाय वन प्लस साइन थीटा इज इक्वल टू सेक थीटा प्लस 10 थीटा का होल स्क्वायर ​

Answer 1

Step-by-step explanation:

Given Question :-

(1 - Sin θ)/(1 + Sin θ) = (Sec θ + Tan θ)^2

Correct Question:-

(1 - Sin θ)/(1 + Sin θ) =( Sec θ -Tan θ)^2

Solution:-

Given that

(1 - Sin θ)/(1 + Sin θ)

On multiplying both numerator and denominator with (1- Sin θ)

=>(1- Sin θ)(1- Sin θ) /(1 + Sin θ)(1 - Sin θ)

=> [(1- Sin θ)]^2 / (1 + Sin θ)(1 - Sin θ)

We know that

(a+b)(a-b)=a^2-b^2

=> [(1- Sin θ)]^2/[(1)^2-(Sin^2 θ)]

=> [(1- Sin θ)]^2/(1-Sin^2 θ)

We know that

Sin^2 A + Cos^2 A = 1

=>>[(1- Sin^2 θ)^2/ Cos^2 θ

=> [(1- Sin θ)/(Cos θ)]^2

=> [(1/Cos θ)-(Sin θ/ Cos θ)]^2

=> (Sec θ - Tan θ)^2

Answer:-

(1 - Sin θ)/(1 + Sin θ) =( Sec θ -Tan θ)^2

Used formulae:-

• a+b)(a-b)=a^2-b^2
• Sin^2 A + Cos^2 A = 1
• Sec A = 1/Cos A
• Tan A = Sin A/Cos A