Question

Plz help class 10 th trigonometry

Answer 1

Trigonometry is not only extremely important for class 10 examinations, but also forms the foundation for higher mathematics. Thus, there are some basic trigonometry formulas for class 10 that are used to derive more complex concepts.

Trigonometry is used to analyze the relationship between angles heights and lengths of triangles. The fundamentals of trigonometry are introduced in class 10 and have been summarized below.
In a right-angled triangle, the Pythagoras theorem states

(perpendicular )2 + ( base )2 = ( hypotenuse )2

There are some properties pertaining to the right-angled triangle. It is to be noted that in the following formulas, P stands for perpendicular, B stands for base and H stands for the hypotenuse.

SinA = P / HCosA = B / HTanA = P / BCotA = B / PCosecA = H / PSecA = H/BSin2A + Cos2A = 1Tan2A + 1 = Sec2A

Cot2A + 1 = Cosec2A

In order to find a relationship between various trigonometric identities, there are some important formulas:
1. TanA = SinA / CosA
2. CotA = CosA / SinA
3. CosecA = 1 / SinA
4. SecA = 1 / CosA

There are some formulas that are very crucial to solving higher level sums.

Sin (A +B) =SinA . CosB + CosA . SinBSin (A – B) = SinA . CosB – CosA . SinBCos (A + B) = CosA . CosB – SinA . SinBCos (A – B) = CosA. CosB + SinA . SinBTan (A + B) = TanA + TanB / 1 – TanA . TanBTan (A – B) = TanA –TanB / 1 + TanA . TanBSin ( A + B) . Sin (A – B) = Sin2A – Sin2B = Cos2B – Cos2ACos (A + B) . Cos (A – B) = Cos2A – Sin2B = Cos2B – Sin2ASin2A = 2 . SinA . CosA = 2 . TanA / (1 + Tan2A)Cos2A = Cos2A – Sin2A = 1 – 2Sin2A = 2Cos2A – 1 = (1 – Tan2A) / (1 + Tan2A)Tan2A = 2TanA / (1 – Tan2A) Sin3A = 3 . SinA – 4 . Sin3ACos3A = 4 . Cos3A – 3 . CosATan3A = (3TanA – Tan3A) / (1 – 3Tan2A)SinA + SinB = 2 Sin (A + B)/2 Cos (A – B)/2SinA – SinB = 2 Sin (A – B)/2 Cos (A + B)/2CosA + CosB = 2 Cos(A – B)/2 Cos (A + B)/2CosA – CosB = 2 Sin(B – A)/2 Sin (A + B)/2TanA + TanB = Sin (A + B) / CosA . CosBSinA CosB = Sin (A + B) + Sin (A – B)CosA SinB = Sin (A + B) – Sin (A – B)CosA CosB = Cos (A + B) + Cos (A – B) SinA SinB = Cos (A – B) – Cos (A + B)

 

Some important questions:
1. If SecA = 13/12, calculate the other ratios:
Answer: SecA = 13/12 = H / BThus,  H = 13, B = 12
Using Pythagoras theorem, we can get the value of P = 5
Sin A = P / H = 5 / 13
Cos A = B / H = 12 / 13
Tan A = P / B = 5 / 12
Cosec A = H / P = 13 / 5
CotA = B / P = 12 / 5Prove that (sec2A – 1) . cot2A = 1 (source: https://www.vedantu.com/rs-aggarwal-solutions/class-10-chapter-8-trigonometric-identities)
Answer: (sec2A – 1) . cot2A
= tan2A x cot2A (As, sec2A – tan2A = 1)
= (1 / cot2A) x cot2A
= 1
= RHSProve that sin2A + 1 / (1 + tan2A) = 1 (source: https://www.vedantu.com/rs-aggarwal-solutions/class-10-chapter-8-trigonometric-identities)
Answer: sin2A + 1 / (1 + tan2A)= sin2A + 1 / (sec2A) (As, 1 + tan2A = sec2A)
= sin2A + cos2A
= 1
= RHSIn triangle PQR, which has a right angle at Q, RP + RQ = 25 inches and QP = 5 inches. What is the value of sinP, tanP and cosP?Answer: We are provided with RP + RQ = 25 inch and QP = 5 inch
Thus, we can say that RP = 25 – RQ
By using the Pythagoras theorem, we get
(QP)2 = (PR)2 – (RQ)2
52  = (25 – RQ)2 – RQ2
25 = 625 + RQ2 – 50 QR – RQ2
25 = 625 – 50 QR
50 QR = 625 – 25 – 600
RQ = 12
Putting the value of RQ in the equation, we get RP = 13.
Therefore, QP = 5 = P
RQ = 12 = B
RP = 13 = H
sinP = P / H = 5 / 13
cos P = B / H = 12 / 13
tan P = P / B = 5 / 12If cos (A + B) sin (C + D) = cos (A – B) sin (C – D), then prove that cotA cotB cotC = cotD (Source: https://schools.aglasem.com/9022)Answer: Since, cos (A + B) sin (C + D) = cos (A – B) sin (C – D),
= cos (A – B) / cos (A – B) = sin (C + D) / sin (C – D)
= cos (A – B) + cos (A + B) / cos (A – B) – cos (A + B) = sin (C + D) + sin (C – D) / sin (C + D) – sin (C – D) (componendo dividendo)
= 2 cosA cosB / 2 sinA sinB = 2 sinC cosD / 2 cosC sinD
= cotA cotB = tanC cotD
= cotA cotB cotC = cotD