Q7 of 54 Page 312

Prove that:

sin (70°+θ) — cos (20° — θ) = 0

Consider L.H.S.

= sin (70°+θ) — cos (20° — θ)


= sin (70°+θ) — cos [90°-(70° + θ)]


= sin (70°+θ) — sin (70° + θ)


= 0 = R.H.S.


Hence, proved.

More from this chapter

All 54 →
6

cos 1° cos 2° cos 3° ... cos 180° = 0

 6

7

tan (55° — θ) — cot (35° + θ) = θ

 7

cosec (67° + θ) — sec (23° — θ) = 0