prove that (sinA +1+ cosA) (sinA-1+ cosA) secA cosecA = 2
Answers
Step-by-step explanation:
LHS=(sinA+1+cosA)(sinA-1+cosA)secA*CosecA. =[(sinA+cosA)+1][sinA+cosA-1]secA*cosecA
=[(sinA+cosA)^2-(1)^2]secA*cosecA
=(sin^2A+2sinA*cosA+cos^2A-1)secA*cosecA
=[(sin^2A+cos^2A)+2sinA*cosA-1]secA*cosecA
=(1+2sinA*cosA-1)secA*cosecA
=2sinA*cosA*secA*cosecA
=2sinA*cosA*(1/sinA)*(1/cosA)
=2=RHSproved
EXPLANATION.
⇒ (sin A + 1 - cos A)(sin A - 1 + cos A)sec A * cosec A = 2.
As we know that,
⇒ secθ = 1/cosθ.
⇒ cosecθ = 1/sinθ.
⇒ sin²θ + cos²θ = 1.
⇒ cos²θ = 1 - sin²θ.
Now, we can write expression as,
⇒ [(sin A + cos A)² - (1)²]sec A x cosec A.
⇒ [sin²A + cos²A + 2sinAcosA - 1]sec A x cosec A.
⇒ [1 + 2sinAcosA - 1]sec A x cosec A.
⇒ [2sinAcosA]sec A x cosec A.
⇒ [2sinAcosA] x 1/(cos A x sin A).
⇒ 2.
Hence proved.
MORE INFORMATION.
Trigonometric ratios of multiple angles.
sin2θ = 2sinθcosθ = 2tanθ/(1 + tan²θ).
cos2θ = 2cos²θ - 1 = 1 - 2sin²θ = cos²θ - sin²θ = (1 - tan²θ)/(1 + tan²θ).
tan2θ = 2tanθ/(1 - tan²θ).
sin3θ = 3sinθ - 4sin³θ.
cos3θ = 4cos³θ - 3cosθ.
tan3θ = (3tanθ - tan³θ)/(1 - 3tan²θ).