证明 (1 – sin A)/(1 + sin A) = (sec A – tan A)²
三角学名称本身表示它是一门处理三角形几何的学科,当需要找到何时给定一些边并且我们需要边之间的关系或边之间的角度时,它非常有用。在三角学中,我们有不同的比率,即 sin A、cos A、tan A、cot A、sec A、cosec A 借助这些比率,可以获得三角形边之间的关系和边之间的角度。
三角函数
三角函数定义了边和角之间的关系,例子有 sin A, cos A, tan A, cot A, sec A, cosec A。不同三角函数之间的关系是三角恒等式。恒等式对于检验三角方程中的不等式非常有用。例子是,
- Tan A = sin A/cos A
- 罪 A = 1/cosec A
- cos A = 1/秒 A
- 棕褐色 A = 1/婴儿床 A
证明 (1 – sin A)/(1 + sin A) = (sec A – tan A)²
为了解决上述问题陈述需要基本恒等式,让我们看一下在这种情况下所需的 6 个三角函数的一些基本恒等式,
证明中使用的先决条件身份
- 罪2 A + cos 2 A = 1
1 – sin 2 A = cos 2 A - tan A = sin A/cos A
- 秒 A = 1/cos A
- 罪 A = 1/cosec A
- (a – b) 2 = a 2 – 2 × a × b + b 2
- a 2 – b 2 = (a + b)(a – b)
- (a – b)/c = a/c – b/c
- sin 2A = 2 × sin A × cos A
给定三角方程
(1 – sin A)/(1 + sin A) = (sec A – tan A) 2
LHS = (1 – sin A)/(1 + sin A)
RHS = (sec A – tan A) 2
从 LHS 端导出证明
Given LHS
(1 – sin A)/(1 + sin A)
Step-1
Multiplying with (1 – sin A)/(1 – sin A) which is equal to 1 to bring the degree of 2 which is present on RHS.
(1 – sin A)(1 – sin A)/(1 + sin A) (1 – sin A)
=(1 – sin A)2/(1 – sin2A)
Step-2
Expanding the numerator (1 – sinA)2
=(1 – 2 × sin A + sin2A)/cos2A
Step-3
Breaking the equation in step-2 in general form
=(1/cos2A) – 2(sin A/cos A)(1/cos A) + (sin2A/cos2A)
Step-4
Substituting with standard formulas in the equation obtained in step-3
=(sec 2A) – 2(tan A) × (sec A) + tan2 A
=(sec A – Tan A)2
从步骤 4 可以得出结论,LHS = (sec A – Tan A) 2等于 RHS,因此,
(秒 A – 棕褐色 A) 2 = (秒 A- 棕褐色 A) 2
左轴 = 右轴
因此证明。
从 RHS 端导出证明
Given RHS
(sec A – Tan A)2
Step-1 Simplifying the equation by substituting standard formulas
=((1/cos A) – ( sin A/cos A))2
=((1 – sin A)/cos A)2
Step 2 Simplifying the denominator
=((1 – sin A)(1 – sin A))/(1 – sin2A)
=((1 – sin A)(1 – sin A))/( (1 – sin A)(1 + sin A))
Step-3
(1 – sin A) in numerator and denominator of the equation in step-2 gets cancelled so it becomes
=(1 – sin A)/(1 + sin A)
从第 3 步可以得出结论,RHS = (1 – sin A)/(1+ sin A) 等于 LHS,因此,
(1 – sin A)/(1 + sin A)=(1 – sin A)/(1 + sin A)
左轴 = 右轴
因此证明。
示例问题
问题1:求解三角恒等式:((cosec A – 1)/(cosec A+1)) × ((1 – sin 2 A)/(1 – sinA) 2 )
解决方案:
- By using identity 4 in the equation
= {((1/sin A) – 1)/((1/sin A) + 1)) × ((1 – sin2A)/(1 – sinA)2}
- By using identity 1 in the equation
= ((1 – sin A)/(1 + sin A)) × ((cosA)2/(1 – sin A)2)
- Multiplying and dividing by cos2A
= ((1 – sin A)/(1 + sin A)) × (1/(sec A – Tan A)2)
- By using our derived identity
= (sec A – Tan A)2 x (1/(sec A – Tan A)2)
= 1
问题 2:求解三角恒等式:((sec A/2) + 2sinA/2)/((sec A/2) – 2sinA/2) ) × (4/(sec A – Tan A) 2 )
解决方案:
- By using identity-3
= 4 × ((1/(cos A/2) + 2sinA/2))/((1/(cos A/2) – 2sinA/2)) × (1/(sec A – Tan A)2)
- By using the identity-8
= 4 × ((1+ 2 sin A/2 × cos A/2)/(1 – 2 sin A/2 × cos A/2)) × (1/(sec A – Tan A)2)
= 4 × ((1 + sin A )/(1 – sin A)) × (1/(sec A – Tan A)2)
- By using the identity proved
= 4 × (sec A – Tan A)2 × (1/(sec A – Tan A)2)
= 4