Question

answer the both questions soon plzzz
1)Prove that.
(Sin A + cosec A)2
+ (cos A + sec A)2
= 7 + tan2 A + cot2 A
2)prove that sin theta- cos theta +1/sin theta +cos theta -1 =1/sec theta - tan theta

Answer 1

hope this helps if so mark as brainliest

Answer 2

         ▶ Method of Solution ◀   ( Sin A + Cosec A ) ² + ( Cos A + Sec A ) ²     ➩ ( Sin² A + Cosec² A + 2 Sin A . Cosec A ) + Cos²A + Sec² A + 2 Cos A. Sec A )    ➩ (Sin² A + Cosec² A + 2 Sin . 1/ Sin ) + Cos²A + Sec² A + 2 Cos A. 1/Cos )     ➩ (Sin² A + Cosec² A + 2 ) + ( Cos²A + Sec² A + 2 )     ➩ ( Sin² A + Cos² A ) + 4 + ( Cosec² A + Sec² A )     ➩ ( 1 + 4 + ( 1+ cot² A ) + ( 1+ tan²A)     ➩  5 + 1 + Cot²A + 1 + tan² A     ➩  5 + 1 + 1 + tan² A + Cot² A     ➩  7 + tan² A + Cot² A    Proved!    (2)     ➩  ( Sin A - Cos A + 1 ) ÷ ( Sin A + Cos A - 1 )     ➩  ( Sin A / Cos A -1 + 1/Cos A ) ÷ ( Sin A/ Cos A + 1 - 1/ Cos A )     ➩  ( tan A - 1 + Sec ) ÷ ( tan A + 1 - Sec )     ➩  ( tan A + Sec - 1 ) ÷ ( tan A - Sec + 1 )     ➩  ( sec A + tan A ) - ( Sec² A - tan² A ) ÷ ( tan A - Sec + 1 )     ➩  Sec A + tan A - [ ( Sec A + tan A ) (SecA - tan A ) ÷ ( tan A - Sec + 1 )     ➩  ( Sec A + tan A ) [ 1 - ( Sec A - tan A ) ÷ ( tan A - Sec + 1 )     ➩ (Sec A + tan A ) [ tan A - Sec A + 1 ] ÷ (tan A - Sec A + 1 )     ➩  ( Sec A + tan A )    Now, RHS    RHS : 1/sec A - tan A     ➩  1/ Sec A - tan A     ➩  1 / Sec - tan A * Sec + tan A / Sec A + tan A     ➩ Sec + tan A ÷ Sec ² A- tan ² A     ➩  Sec A + tan A ÷ 1     ➩ Sec A + tan A                  Proved !