tan(t)=1cot(t)cot(t)=1tan(t)csc(t)=1sin(t)sin(t)=1csc(t) Answer and Explanation:1 Given Data: The given identity is:tanxcotxsinx=cscx. Simplify the left-hand side of the given identity. ...
可得到下列恒等式（identity）：tanθ＝sinθ/cosθ,cotθ＝cosθ/sinθsecθ＝1/cosθ,cscθ＝1/sinθ分别用cos 2θ与sin 2θ来除cos 2θ＋sin 2θ＝1,可得：sec 2θ–tan 2θ＝1 及 csc 2θ–cot 2θ＝1对于负角度,六个三角函数分别为：sin(–θ)＝ –sinθ csc(–θ)＝ –cs...
Given: {eq}\dfrac{\tan(x)\cot(x)}{\cos(x)}=\sec(x) {/eq} Our objective is to verify identity: First, we must know that the cotangent is the...Become a member and unlock all Study Answers Start today. Try it now Create an account Ask a question Our experts...
View Solution The values of θ lying between 0 and π/2 and satisfying the equation∣∣1+sin2θsin2θsin2θcos2θ1+cos2θcos2θ4sin4θ4sin4θ1+4sin4θ∣∣=0 are View Solution Free Ncert Solutions English Medium NCERT Solutions ...
根据这些定义,可得到下列恒等式(identity):tanθ=sinθ/cosθ,cotθ=cosθ/sinθ secθ=1/cosθ,cscθ=1/sinθ 分别用cos 2θ与sin 2θ来除cos 2θ+sin 2θ=1,可得:sec 2θ–tan 2θ=1 及 csc 2θ–cot 2θ=1 对于负角度,六个三角函数分别为:sin(–θ)=–sinθ csc(–θ)=–cscθ cos...
Step 4: Simplify further using double angle identityWe can use the double angle identity for sine:sin(θ)=2sin(θ2)cos(θ2)So, we can rewrite the expression as:=2sin(θ) Step 5: Final resultThus, the simplified value of tan(θ2)+cot(θ2) is:=2⋅1sin(θ)=2sin(θ) Show Mor...
数学中tan是正切的意思。角θ在任意直角三角形中，与θ相对应的对边与邻边的比值叫做角θ的正切值。若将θ放在直角坐标系中即tanθ=y/x。tanA=对边/邻边。在直角坐标系中相当于直线的斜率k。
How do you verify the identitycotα+tanα=cscαsecα? https://socratic.org/questions/how-do-you-verify-the-identity-cotalpha-tanalpha-cscalphasecalpha See proof below Explanation:cos2α+sin2α=1cotα=sinαcosαtanα=cosαsinα... ...
根据这些定义，可得到下列恒等式（identity）：tanθ＝sinθ/cosθ，cotθ＝cosθ/sinθ secθ＝1/cosθ，cscθ＝1/sinθ 分别用cos 2θ与sin 2θ来除cos 2θ＋sin 2θ＝1，可得：sec 2θ–tan 2θ＝1 及 csc 2θ–cot 2θ＝1 对于负角度，六个三角函数分别为：sin(–θ)＝ –sin...
Verify the trigonometric identity: sin x (tan x + 1 / {tan x}) = sec x. Verify the trigonometric identity: {sin^2 x} / {1 - cos x} = 1 + cos x. Verify the trigonometric identity. \dfrac{tan(x)cot(x)}{cos(x)}=sec(x) Verify the trigonometric identity. \frac{\s...