Problem & Solution -
cos A sin A
Example.1) Prove that, ------------ + ------------ = (cos A + sin A)
(1 – tan A) (1 – cot A)
cos A sin A
Ans.) L.H.S = ------------- + -------------
(1 – tan A) (1 – cot A)
cos A sin A
= ------------------ + -----------------
sin A cos A
(1 - --------) (1 - --------)
cos A sin A
cos A sin A
= -------------------- + ----------------------
cos A - sin A sin A - cos A
(----------------) (----------------)
cos A sin A
cos² A sin² A
= --------------- + ----------------
cos A - sin A sin A - cos A
sin² A cos² A
= ----------------- - ----------------
sin A - cos A sin A - cos A
(sin² A - cos² A)
= -------------------
(sin A - cos A)
(sin A + cos A) (sin A + cos A)
= -------------------------------
(sin A - cos A)
= sin A + cos A = R.H.S
Hence, L.H.S = R.H.S (Proven)
(1 + sin A) cos A
Example.2) Prove that, √------------- = ------------
(1 – sin A) (1 – sin A)
(1 + sin A)
Ans.) We have, L.H.S = √--------------
(1 – sin A)
√(1 + sin A)
= ----------------
√(1 – sin A)
multiplying numerator & denominator by √(1 – sin A), and we get -
√(1 + sin A) √(1 – sin A)
= --------------- X ---------------
√(1 – sin A) √(1 – sin A)
√(1 - sin² A)
= -----------------
(1 – sin A)
√cos² θ
= -------------
(1 – sin A)
cos θ
= ----------- = R.H.S
(1 – sin A)
So, L.H.S = R.H.S
(1 + sin A) cos A
√------------- = ------------ (Proven)
(1 – sin A) (1 – sin A)
tan A + sec A - 1 1 + sin A
Example.3) Prove that, --------------------- = --------------
tan A – sec A + 1 cos A
tan A + sec A - 1
Ans.) We have, L.H.S = ---------------------
tan A – sec A + 1
[substitute 1 = sec² A - tan² A]
tan A + sec A – (sec² A - tan² A)
= -------------------------------------
tan A – sec A + 1
(tan A + sec A) – {(sec A + tan A) (sec A - tan A)}
= -----------------------------------------------
tan A – sec A + 1
(tan A + sec A) – {(sec A + tan A) (sec A - tan A)}
= ------------------------------------------------
tan A – sec A + 1
(tan A + sec A) {1 – (sec A - tan A)}
= ------------------------------------
tan A – sec A + 1
(tan A + sec A) (1 + tan A – sec A)
= -----------------------------------
(1 + tan A – sec A)
= (tan A + sec A)
sin A 1
= (---------- + -------)
cos A cos A
(1 + sin A)
= ------------- = R.H.S
cos A
So, L.H.S = R.H.S
tan A + sec A - 1 1 + sin A
Hence, --------------------- = ------------- (Proved)
tan A – sec A + 1 cos A
Example.4) If x = a sin θ + b cos θ and y = a cos θ – b sin θ, prove that : x² + y² = a² + b²
Ans.) We have –
L.H.S = x² + y² = (a sin θ + b cos θ)² + (a cos θ – b sin θ)² [replacing the value of x & y]
= a² sin² θ + b² cos² θ + 2ab sin θ. Cos θ + a² cos² θ + b² sin² θ – 2ab sin θ. Cos θ
= a² (sin² θ + cos² θ) + b² (cos² θ + sin² θ)
= a².1 + b².1 [as we know, sin² θ + cos² θ = 1]
= a² + b² = R.H.S
So, x² + y² = a² + b² (Proved)
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