If cos ( α - β ) + cos ( β - γ ) + cos ( γ - α ) = - 3 2 , then prove that cos α + cos β + cos γ = sin α + sin β + sin γ = 0 .

Given that cos ( α - β ) + cos ( β - γ ) + cos ( γ - α ) = - 3 2 ⇒ cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α = - 3 2 ⇒ 2 cos α cos β + 2 sin α sin β + 2 cos β cos γ + 2 sin β sin γ + 2 cos γ cos α + 2 sin α sin γ = - sin 2 α + cos 2 α + sin 2 β + cos 2 β + sin 2 γ + cos 2 γ ⇒ sin 2 α + sin 2 β + sin 2 γ + 2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α + cos 2 α + cos 2 β + cos 2 γ + 2 cos α cos β + 2 cos β cos γ + 2 cos α cos γ = 0 ⇒ sin α + sin β + sin γ 2 + cos α + cos β + cos γ 2 = 0 ⇒ ( sin α + sin β + sin γ ) = cos α + cos β + cos γ = 0 Hence Proved.

Show that tan 45 ° + A = 1 + tan A 1 - tan A .

To Prove : tan 45 ° + A = 1 + tan A 1 - tan A Given that L . H . S = tan 45 ° + A = tan 45 ° + tan A 1 - tan 45 ° tan A = 1 + tan A 1 - tan A = R . H . S Hence Proved.

Show that tan 45 ° - A = 1 - tan A 1 + tan A .

To Prove : tan 45 ° - A = 1 - tan A 1 + tan A Given that L . H . S = tan 45 ° - A = tan 45 ° - tan A 1 + tan 45 ° tan A = 1 - tan A 1 + tan A = R . H . S Hence Proved.

Prove that cot ( A + B ) = cot A cot B - 1 cot A + cot B .

To Prove : cot ( A + B ) = cot A cot B - 1 cot A + cot B . Given that L . H . S = cot ( A + B ) = cos ( A + B ) sin ( A + B ) = cos A cos B - sin A sin B sin A cos B + cos A sin B Divide numerator and denominator by sin A sin B = cos A cos B sin A sin B - 1 sin A cos B sin A sin B + cos A sin B sin A sin B = cot A cot B - 1 cot B + cot A = cot A cot B - 1 cot A + cot B = R . H . S Hence Proved.

Prove that tan π 4 + θ tan 3 π 4 + θ = - 1 .

To Prove : tan π 4 + θ tan 3 π 4 + θ = - 1 Given that L . H . S = tan π 4 + θ tan 3 π 4 + θ We know that tan ( A + B ) = tan A + tan B 1 - tan A tan B = tan π 4 + tan θ 1 - tan π 4 tan θ · tan 3 π 4 + tan θ 1 - tan 3 π 4 · tan θ We know that tan ( 180 - θ ) = - tan θ = 1 + tan θ 1 - tan θ tan θ + tan π - π 4 1 - tan π - π 4 · tan θ = 1 + tan θ 1 - tan θ tan θ - tan π 4 1 - - tan π 4 · tan θ = 1 + tan θ 1 - tan θ tan θ - 1 ) 1 + tan θ = - 1 = R . H . S Hence Proved.