Exercise 6.3 Solutions
Step-by-step solved textbook exercises verifying fundamental trigonometric identities and computing remaining ratios given one key ratio in the first quadrant.
Since θ is in the first quadrant, all six trigonometric ratios are positive. We represent a right-angled triangle with perpendicular a, base c, and hypotenuse b (where b2 = a2 + c2).
(i) Given sin θ = 23
Here, perpendicular a = 2 and hypotenuse b = 3.
By Pythagoras' Theorem: b2 = a2 + c2
32 = 22 + c2
9 = 4 + c2 ⇒ c2 = 5 ⇒ c = √5
Thus, the remaining ratios are:
(ii) Given cos θ = 34
Here, base c = 3 and hypotenuse b = 4.
By Pythagoras' Theorem: b2 = a2 + c2
42 = a2 + 32
16 = a2 + 9 ⇒ a2 = 7 ⇒ a = √7
Thus, the remaining ratios are:
(iii) Given tan θ = 12
Here, perpendicular a = 1 and base c = 2.
By Pythagoras' Theorem: b2 = a2 + c2
b2 = 12 + 22 = 1 + 4 = 5 ⇒ b = √5
Thus, the remaining ratios are:
(iv) Given sec θ = 3
Since sec θ = 3 = 31, we have hypotenuse b = 3 and base c = 1.
By Pythagoras' Theorem: b2 = a2 + c2
32 = a2 + 12
9 = a2 + 1 ⇒ a2 = 8 ⇒ a = √8 = 2√2
Thus, the remaining ratios are:
(v) Given cot θ = √3√2
Here, base c = √3 and perpendicular a = √2.
By Pythagoras' Theorem: b2 = a2 + c2
b2 = (√2)2 + (√3)2 = 2 + 3 = 5 ⇒ b = √5
Thus, the remaining ratios are:
L.H.S = (sin θ + cos θ)2
Using the algebraic expansion formula: (x + y)2 = x2 + y2 + 2xy
L.H.S = sin2 θ + cos2 θ + 2 sin θ cos θ
Since we know the fundamental identity: sin2 θ + cos2 θ = 1
L.H.S = 1 + 2 sin θ cos θ = R.H.S
Hence Proved.
L.H.S = cos θsin θ tan θ
Recall that tan θ = sin θcos θ:
Since cot θ = cos θsin θ:
L.H.S = (cos θsin θ)2 = cot2 θ = R.H.S
cot θ, but then correctly concludes L.H.S = R.H.S which is cot2 θ. We have written the mathematically rigorous correction here showing the square.Hence Proved.
L.H.S = sin θcosec θ + cos θsec θ
Using the reciprocal identities cosec θ = 1sin θ and sec θ = 1cos θ:
L.H.S = (sin θ × sin θ) + (cos θ × cos θ)
L.H.S = sin2 θ + cos2 θ
Since sin2 θ + cos2 θ = 1:
L.H.S = 1 = R.H.S
Hence Proved.
L.H.S = cos2 θ − sin2 θ
Using the fundamental identity sin2 θ = 1 − cos2 θ:
L.H.S = cos2 θ − 1 + cos2 θ
L.H.S = 2 cos2 θ − 1 = R.H.S
Hence Proved.
L.H.S = cos2 θ − sin2 θ
Using the fundamental identity cos2 θ = 1 − sin2 θ:
L.H.S = 1 − 2 sin2 θ = R.H.S
Hence Proved.
L.H.S = 1 − sin θcos θ
To establish the identity, we multiply both numerator and denominator by the conjugate of the numerator, which is (1 + sin θ):
L.H.S = (1 − sin θ)(1 + sin θ)cos θ(1 + sin θ)
L.H.S = 1 − sin2 θcos θ(1 + sin θ)
Using the identity 1 − sin2 θ = cos2 θ:
L.H.S = cos θ1 + sin θ = R.H.S
Hence Proved.
L.H.S = (sec θ − tan θ)2
Expressing sec θ and tan θ in terms of sin and cos:
L.H.S = (1 − sin θcos θ)2 = (1 − sin θ)2cos2 θ
Substitute cos2 θ = 1 − sin2 θ:
Expanding the denominator as a difference of squares: 1 − sin2 θ = (1 − sin θ)(1 + sin θ):
L.H.S = 1 − sin θ1 + sin θ = R.H.S
Hence Proved.
L.H.S = (tan θ + cot θ)2
Expressing tan θ and cot θ in terms of sin and cos:
L.H.S = (sin2 θ + cos2 θcos θ sin θ)2
Substitute sin2 θ + cos2 θ = 1:
L.H.S = 1cos2 θ sin2 θ
L.H.S = 1cos2 θ × 1sin2 θ
L.H.S = sec2 θ cosec2 θ = R.H.S
Hence Proved.
L.H.S = tan θ + sec θ − 1tan θ − sec θ + 1
Recall the trigonometric identity: 1 + tan2 θ = sec2 θ which gives:
1 = sec2 θ − tan2 θ
Substitute this expression for the constant "1" in the numerator:
Factor the difference of squares: sec2 θ − tan2 θ = (sec θ + tan θ)(sec θ − tan θ):
Factor out the common term (tan θ + sec θ) from the numerator:
L.H.S = (tan θ + sec θ)[1 − sec θ + tan θ]tan θ − sec θ + 1
Observe that [1 − sec θ + tan θ] is identical to [tan θ − sec θ + 1] in the denominator. Canceling these common factors yields:
L.H.S = tan θ + sec θ = R.H.S
Hence Proved.
L.H.S = sin3 θ − cos3 θ
Using the algebraic identity: x3 − y3 = (x − y)(x2 + xy + y2)
Grouping the squared terms: sin2 θ + cos2 θ = 1:
L.H.S = (sin θ − cos θ)(1 + sin θ cos θ) = R.H.S
Hence Proved.
L.H.S = sin6 θ − cos6 θ
Write the expression as a difference of cubes: (x2)3 − (y2)3
Applying the formula A3 − B3 = (A − B)(A2 + AB + B2), where A = sin2 θ and B = cos2 θ:
Now, let's simplify the bracket term: (sin2 θ)2 + (cos2 θ)2 + sin2 θ cos2 θ.
Recall that x2 + y2 = (x + y)2 − 2xy. Setting x = sin2 θ and y = cos2 θ:
Since sin2 θ + cos2 θ = 1:
(sin2 θ)2 + (cos2 θ)2 = (1)2 − 2 sin2 θ cos2 θ = 1 − 2 sin2 θ cos2 θ
Substitute this back into the bracket term:
L.H.S = (sin2 θ − cos2 θ)(1 − sin2 θ cos2 θ) = R.H.S
Hence Proved.
🎲 Exercise 6.3 CAST Rule & Identity Solver
Select any trigonometric identity and slide the angle from 0° to 360° to dynamically evaluate LHS & RHS, check the quadrant, and verify signs using the CAST rule!