The correct answer is option 2 i.e. tan t. sec t.
Given that
4−4sint1+sint−4+4sint1−sint
⇒ (4−4sint)(4+4sint)(1+sint)(4+4sint)−(1−sint)(4−4sint)
⇒ 16[1−(sint)(sint)]4+4sint+4sint+(4sint)(sint)−4+4sint+4sint−(4sint)(sint)
⇒ 16(cost)(cost)16sint
⇒ tant. sect