2015年9月21日 · Simplify: csc^2 x - 1 Ans: cot^2 x 1/sin^2 x - 1 = (1 - sin^2 x)/sin^2 x = cos^2 x /sin^2 x = cot^2 x

2015年12月13日 · Prove 1 + cot^2 x = csc^2 x 1 + cot^2 x = 1 + cos^2 x/(sin^2 x) = (sin^2 x + cos^2 x)/(sin^2 x) = 1/(sin^2 x) = csc^2 x

2018年4月13日 · #csc^2(x)=1/sin^2(x)# #d/dx[csc^2(x)]=d/dx[1/sin^2(x)]# #d/dx[1/sin^2(x)]=d/dx[[sin(x)]^{-2}]# let #u=sinx#. #d/dx[[sin(x)]^{-2}]=d/{du}[u^{-2}]d/dx[sinx]#

2015年3月15日 · The antiderivative of csc^2x is -cotx+C. Why? Before trying anything 'fancy' (subsitutuion, parts, trig sub, misc sub, partial fractions, et c.) try 'staightaway' antidifferentiation. Do you know a finction whose derivative is csc^2x? Go through the list: d/(dx)(sinx)=cosx d/(dx)(cosx)=-sinx d/(dx)(tanx)=sec^2x Hang on! that's good! The derivative if a co function …

2016年11月23日 · csc^2x-1 is equal to cot^2x Use the identity: sin^2x+cos^2x=1 Divide each term in the identity by sin^2theta 1+cot^2x=csc^2x Subtract 1 from each side: cot^2x=csc^2x-1

2018年4月25日 · 5651 views around the world You can reuse this answer ...

2015年8月5日 · How do you apply the fundamental identities to values of #theta# and show that they are true

2016年5月4日 · Note the following identity: #csc^2theta = 1 + cot^2theta# So, let's see how that works out. #\mathbf(sec^2theta(csc^2theta) = sec^2theta + csc^2theta)#

2016年5月4日 · see below Use Property: cot^2x+1=csc^2x Left Side: = csc^2x-cot^2x =cot^2x+1-cot^2x =cot^2x-cot^2x + 1 =1 = Right Side

2016年5月4日 · see below Left Side:=csc^2 theta (1-cos^2 theta) = csc^2 theta * sin^2 theta = 1/sin^2 theta * sin^2 theta =1 = Right Side