Join 60,000+ competitive exam aspirants
Which of the following options gives the correct expression for y, when 1−sinAcos2A = y, and sinA=1?
\csc A
\sec A + tanA
sinA+1
cosA+1
sinA+1
Use the algebraic identity 1−sin2A=(1−sinA)(1+sinA) to cancel out the denominator directly in 5 seconds.
y=\frac{\cos² A}{1 - \sin A} and sinA=1
cos2A=1−sin2A
Use the algebraic identity 1−sin2A=(1−sinA)(1+sinA) to cancel out the denominator directly in 5 seconds.
Forgetting that cosA1=secA and cosAsinA=tanA, leading to wrong options like sinA+1.
Step 1: Write the given expression
We are given the expression for y: y=1−sinAcos2A
y=1−sinAcos2A
Step 2: Apply trigonometric identity for numerator
Using the fundamental trigonometric identity cos2A=1−sin2A, substitute it into the numerator: y=1−sinA1−sin2A
y=1−sinA1−sin2A
Step 3: Factorize the numerator
The numerator is in the form a2−b2=(a−b)(a+b) where a=1 and b=sinA: y=1−sinA(1−sinA)(1+sinA)
y=1−sinA(1−sinA)(1+sinA)
Step 4: Cancel common terms and simplify
Since sinA=1, 1−sinA=0, so we can cancel 1−sinA from both numerator and denominator: y=1+sinA
y=1+sinA
C is correct because simplifying the expression using cos2A=1−sin2A and factorizing gives 1+sinA.
Similar algebraic factorization and identity substitution techniques are frequently used to simplify complex trigonometric proofs and height-distance problems.