18.(12分)证明下列三角恒等式:18.(12分)证明下列三角恒等式: \$\frac { 1 - \sin 2 \theta } { 1 + \cos 2 \theta } = \frac { ( \tan \theta - 1 ) ^ { 2 } } { 2 }\$ 18.(12分)证明下列三角恒等式: 相关知识点: 试题来源: 解析 18.【解】(1左)边 _ 18.【...
Verify the following identity: (cos(theta) + sin(theta))/(cos(theta) - sin(theta)) - (cos(theta) - sin(theta))/(cos(theta) + sin(theta)) = 2 tan(2 theta). Verify the identity: 1 - (cos^2 x)/(1 - sin x) = -sin x. Verify the Identity: cos^2 t/sin t ...
【题目】若$$ \tan \theta = 2 $$,则$$ \frac { \cos \theta ( 1 - \sin 2 \theta ) } { \sin \theta - \cos \theta } = ( ) $$ A.-$$ \frac { 1 } { 5 } $$ B.$$ \frac { 1 } { 5 } $$ C.$$ \frac { 2 } { 5 } $$ D.$$ \frac { 6 } ...
1−sin2(θ)1-sin2(θ)Because the two sides have been shown to be equivalent, the equation is an identity. cot(θ)sin(θ)cos(θ)=1−sin2(θ)cot(θ)sin(θ)cos(θ)=1-sin2(θ) is an identitycot(θ)sin(θ)cos(θ)=1−sin2(θ)cot(θ)sin(θ)cos(θ)=1-sin...
यदि theta=30^(@)हो तो (1-sin^(2)2theta)/(cos2theta)का मान है- 02:22 cot30^(@) का मान है- 01:05 sec45^(@) का मान है- 00:57 sin^(2)45^(@)का मान है- 01:02 cot^(2)60^(...
sin (180-theta)sin (90- theta)+[cot(90- theta)// 1+tan^2 theta] का मान क्या है? View Solution If tan^4 theta+ tan^2 theta=1 then the value of cos^4 theta+cos^2 theta is यदि tan^4 theta+ tan^2 theta=1 हो, तो cos^4 ...
【答案】D $$ \frac { \sin \theta ( 1 - \sin 2 \theta ) } { \sin \theta - \cos \theta } = \frac { \sin \theta ( \sin \theta - \cos \theta ) } { 1 } = \frac { \sin ^ { 2 } \theta - \sin \theta \cos \theta } { \sin ^ { 2 } \theta + ...
Answer to: Find the slope of the tangent line to the polar curve r at \theta = 0. r = 1 - 2 \sin \theta By signing up, you'll get thousands of...
解法1 $$ \overline { z } = 1 - \sin \theta - i \cos \theta \\ = 1 + \cos ( \frac { 3 \pi } { 2 } - \theta ) + i \sin ( \frac { 3 \pi } { 2 } - \theta ) \\ = 2 \cos ^ { 2 } ( \frac { 3 \pi } { 4 } - \frac { \theta } { 2...
यदिsecθ(cosθ+sinθ)=√2है,तो(2sinθ)/(cosθ−sinθ)का मान क्या है? If sec theta + tan theta = m (>1) , then the value of sin theta is (0^@ यदि sec theta + tan theta = m (>1) हो, तो sin theta ...