Question

Sec power 4 theta minus tan ka power 4 theta equals to 6 square theta + Tan square theta

Answer 1

Step-by-step explanation:

We have given that,

                          $Sec^{4} - Tan^{4} = Sec^{2} + Tan^{2}$            (i)

Now, we know an identity that,

                        $(a + b)(a - b) = (a^{2} -b^{2})$;           (ii)

Hence, from equation (i) and (ii) we have,

               $Sec^{4} - Tan^{4}$  = $(Sec^{2} - Tan^{2} )(Sec^{2} + Tan^{2} )$         (iii)

Now, put value from equation (iii) to (i);

                $(Sec^{2} - Tan^{2} )(Sec^{2} + Tan^{2} )$  = $Sec^{2} + Tan^{2}$ ;           (iv)

From trigonometric identity we know that ,

                 $Sec^{2} - Tan^{2}$ = 1 ;                (v)

Now, put the value of equation (v) in equation (iv);

Hence,                $(1 )(Sec^{2} + Tan^{2} )$  = $Sec^{2} + Tan^{2}$ ;

                          $Sec^{2} + Tan^{2}$  = $Sec^{2} + Tan^{2}$ ;

Hence, the LHS is equal to the RHS.

That's all