If cos 4 ⁡ θ sec 2 ⁡ α , 1 2 , sin 4 ⁡ θ csc 2 ⁡ α are in A.P ,then prove that cos 8 ⁡ θ sec 6 ⁡ α , 1 2 , sin 8 ⁡ θ csc 6 ⁡ α are in A.PNow i have reached upto 1 = cos 4 ⁡ θ sec 2 ⁡ α + sin 4 ⁡ θ csc 2 ⁡ α. By completing square i have ( sin 2 ⁡ θ cot ⁡ α − cos 2 ⁡ θ tan ⁡ α ) 2 = 0 so i get tan ⁡ θ = ± tan ⁡ α. How do i proceed?

Question

If       cos    4    ⁡  θ      sec    2    ⁡  α  ,      1    2    ,      sin    4    ⁡  θ      csc    2    ⁡  α are in A.P ,then prove that       cos    8    ⁡  θ      sec    6    ⁡  α  ,      1    2    ,      sin    8    ⁡  θ      csc    6    ⁡  α are in A.PNow i have reached upto  1  =      cos    4    ⁡  θ      sec    2    ⁡  α  +      sin    4    ⁡  θ      csc    2    ⁡  α. By completing square i have   (      sin    2    ⁡  θ  cot  ⁡  α  −      cos    2    ⁡  θ  tan  ⁡  α      )    2    =  0 so i get   tan  ⁡  θ  =  ±  tan  ⁡  α. How do i proceed?

Anika Stevenson · Accepted Answer

You have done rightly. Now we can proceed as follows:                    (1)                    tan        ⁡        θ        =        ±        tan        ⁡        α        ⇒                  tan          2                ⁡        θ        =                  tan          2                ⁡        α        ⇒                  sec          2                ⁡        θ        −        1        =                  sec          2                ⁡        α        −        1        ⇒                  sec          2                ⁡        θ        =                  sec          2                ⁡        α        ⇔                  sec          6                ⁡        θ        =                  sec          6                ⁡        α            Also,                    (2)                              tan          2                ⁡        θ        =                  tan          2                ⁡        α        ⇒                  cot          2                ⁡        θ        =                  cot          2                ⁡        α        ⇒                  csc          2                ⁡        θ        −        1        =                  csc          2                ⁡        α        −        1        ⇒                  csc          2                ⁡        θ        =                  csc          2                ⁡        α        ⇔                  csc          6                ⁡        θ        =                  csc          6                ⁡        α            Hope you can take it from here by using (1) and (2).

If cos 4 </msup> ⁡ θ sec 2 </msup>

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