if x1, x2, x3......xn in AP then prove thatsecx1.secx2 + secx2.secx3 + secx3.secx4................................secxn-1.secxn = (tanxn-tanx1)/sin(x2-x1)

Question

Ritesh Khatri · Answer

secx r .secx r+1 =1/(cosx r .cosx r+1 ) = sind/(sind.cosx r .cosx r+1) (where d is difference of given AP) =sin(x r+1 -x r )/(sind.cosx r .cosx r+1 ) = (sinx r+1 cosx r – cosx r+1 sinx r )/(sind.cosx r .cosx r+1 ) =[(sinx r+1 cosx r )/(sind.cosx r .cosx r+1 )] – [(cosx r+1 sinx r )/(sind.cosx r .cosx r+1 )] =[(sinx r+1 )/(sind.cosx r+1 )] – [(sinx r )/(sind.cosx r .)] =[tanx r+1 – tanx r ]/sind Now substitute r=1,2,3,....,n-1 and add, we will get (tanx n – tanx 1 )/sind =(tanx n – tanx 1 )/sin(x 2 – x 1 )

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if x1, x2, x3......xn in AP then prove thatsecx1.secx2 + secx2.secx3 + secx3.secx4................................secxn-1.secxn = (tanxn-tanx1)/sin(x2-x1)

if x1, x2, x3......xn in AP then prove thatsecx1.secx2 + secx2.secx3 + secx3.secx4................................secxn-1.secxn = (tanxn-tanx1)/sin(x2-x1)

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