Prove that:sinθ(1+tanθ)+cosθ(1+cotθ)=secθ+cosecθ View Solution Prove that 1(secθ−tanθ)−1cosθ=1cosθ−1(secθ+tanθ). View Solution View Solution Iftanθ=1−m2then prove thatsecθ+tan3θcosecθ=(2−m2)32 ...
sec theta - tan theta = x, then sec theta = ? 03:15 (tan theta+sec theta-1)/(tan theta-sec theta+1) 03:07 The locus of the point sec theta+tan theta,sec theta-tan theta is 01:31 (sec theta+tan theta)/(sec theta-tan theta)+(sec theta-tan theta)/(sec... 02:13 The va...
sec(θ)=32sec(θ)=32,tan(θ)<0tan(θ)<0 The tangent function is negative in the second and fourth quadrants. The secant function is positive in the first and fourth quadrants. The set of solutions forθθare limited to the fourth quadrant since that is the only quadrant found in both...
( (sec)(θ )+C-3(e^(θ )+C))Simplify.( (sec)(θ )-3e^(θ )+C)The answer is the antiderivative of the function( R(θ )=(sec)(θ )(tan)(θ )-3e^(θ )).( 2(θ )=)( (sec)(θ )-3e^(θ )+C)反馈 收藏
tan(θ)=对边相邻tan(θ)=对边相邻 求单位圆三角形的斜边。由于已知相对边和相邻边,所以可以使用勾股定理求第三条边。 斜边=√对边2+相邻2斜边=对边2+相邻2 替换方程中的已知值。 斜边=√(−3)2+(5)2斜边=(-3)2+(5)2 化简根式内部。
Answer to: Verify that the equation is an identity. Show that (sec theta + tan theta)^2 = (1 + sin theta)/(1 - sin theta). By signing up, you'll...
We will use the substitution t=secθ−tanθ Answer and Explanation: Using the substitution, we get: {eq}dt = -\sec \theta (\sec \theta - \tan \theta) d\theta \ I = -\int \frac{\sec \theta (\sec \theta - \tan...Become...
tan(θ)sec(θ) + 2tan(θ) = 0 Begin by factoring out tan(θ): tan(θ)[sec(θ) + 2] = 0 Now, using the Zero Product Rule, set each factor equal to zero: 1) tan(θ) = 0 which occurs at θ = 0 and θ = π 2) sec(θ) + 2 = 0 sec(θ) = -2 1/cos(θ) = ...
y=log(sec\theta+tan\theta)and x=sec\theta find dy/dx Solution Step 2 Unlock Answer Unlock Previous questionNext question Not the question you’re looking for? Post any question and get expert help quickly.
Answer to: Simplify: 1) \frac{-\sin(-x) \cos x \sec x \csc x \tan x}{\cot x} \\2) \frac{1 + \tan^2 \theta}{\csc^2 \theta} + \sin^2 \theta +...