1) If sin (A + B) = 1 and tan (A – B) = 1/√3, find the value of:
tan A + cot B
sec A – cosec B
2) If cosec θ = 5/3, then what is the value of cos θ + tanθ
3) ΔRPQ is a right angled at Q. If PQ = 5 cm and RQ = 10 cm, find:
sin²P
cos²R and tan R
sin P x cos P
sin²P – cos²P
4) If tan(A – B) = 1/√3 and tan (A + B) = √3, find A and B
5) If sec2 θ(1 + sin θ) (1 – sin θ) = k, then find the value of k.
6) If cosec (A-B) = 2, cot (A + B) = 1/√3 ; 0° < (A + B) <90°, A > B, then find A and B
7) ΔABC is right angled at B, BC = 7 cm and AC – AB = 1 cm. Find the value of cos A + sin A
8) Determine the value of x such that 2 cosec² 30° + x sin² 60° -3/4 tan² 30° = 10.
9) If sec A = 15/7 and A + B = 90°, find the value of cosec B
10) If 7x = cosecθ and 7/x = cot θ, find the value of (x²-1/x²)
11) If 6x = sec θ and 6/x = tanθ, find the value of 9(x²-1/x²)
12) If 8x = cosec A and 8/x = cot A, find the value of 4(x²-1/x²)
13) If 4x = sec θ and 4/x = tan θ, find the value of 8(x²-1/x²)
Answers
Answer:
= cosec A + cot A, using the identity cosec2A = 1+cot2A.
With the help of identity function, cosec2A = 1+cot2A, let us prove the above equation.
L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)
Divide the numerator and denominator by sin A, we get
= (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A
We know that cos A/sin A = cot A and 1/sin A = cosec A
= (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)
= (cot A – cosec2A + cot2A + cosec A)/(cot A+ 1 – cosec A) (using cosec2A – cot2A = 1
= [(cot A + cosec A) – (cosec2A – cot2A)]/(cot A+ 1 – cosec A)
= [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)
= (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
= cot A + cosec A = R.H.S.
Therefore, (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A
Hence Proved
Ncert solutions class 10 chapter 8-11
First divide the numerator and denominator of L.H.S. by cos A,
Ncert solutions class 10 chapter 8-12
We know that 1/cos A = sec A and sin A/ cos A = tan A and it becomes,
= √(sec A+ tan A)/(sec A-tan A)
Now using rationalization, we get
Ncert solutions class 10 chapter 8-13
= (sec A + tan A)/1
= sec A + tan A = R.H.S
Hence proved
(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
L.H.S. = (sin θ – 2sin3θ)/(2cos3θ – cos θ)
Take sin θ as in numerator and cos θ in denominator as outside, it becomes
= [sin θ(1 – 2sin2θ)]/[cos θ(2cos2θ- 1)]
We know that sin2θ = 1-cos2θ
= sin θ[1 – 2(1-cos2θ)]/[cos θ(2cos2θ -1)]
= [sin θ(2cos2θ -1)]/[cos θ(2cos2θ -1)]
= tan θ = R.H.S.
Hence proved
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
It is of the form (a+b)2, expand it
(a+b)2 =a2 + b2 +2ab
= (sin2A + cosec2A + 2 sin A cosec A) + (cos2A + sec2A + 2 cos A sec A)
= (sin2A + cos2A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan2A + 1 + cot2A
= 1 + 2 + 2 + 2 + tan2A + cot2A
= 7+tan2A+cot2A = R.H.S.
Therefore, (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A
Hence proved.
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
First, find the simplified form of L.H.S
L.H.S. = (cosec A – sin A)(sec A – cos A)
Now, substitute the inverse and equivalent trigonometric ratio forms
= (1/sin A – sin A)(1/cos A – cos A)
= [(1-sin2A)/sin A][(1-cos2A)/cos A]
= (cos2A/sin A)×(sin2A/cos A)
= cos A sin A
Now, simplify the R.H.S
R.H.S. = 1/(tan A+cotA)
= 1/(sin A/cos A +cos A/sin A)
= 1/[(sin2A+cos2A)/sin A cos A]
= cos A sin A
L.H.S. = R.H.S.
(cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
Hence proved
(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A
L.H.S. = (1+tan2A/1+cot2A)
Since cot function is the inverse of tan function,
= (1+tan2A/1+1/tan2A)
= 1+tan2A/[(1+tan2A)/tan2A]
Now cancel the 1+tan2A terms, we get
= tan2A
(1+tan2A/1+cot2A) = tan2A
Similarly,
(1-tan A/1-cot A)2 = tan2A
Hence proved