to download NCERT Solutions for Class 10 Maths chapter 8 Introduction to Trigonometry, Ans : \(\begin{array}{l}{\text { Applying Pythagoras theorem for } \Delta A B C, \text { we obtain }} \\ {A C^{2}=A B^{2}+B C^{2}} \\ {=(24 \mathrm{cm})^{2}+(7 \mathrm{cm})^{2}} \\ {=(576+49) \mathrm{cm}^{2}} \\ {=625 \mathrm{cm}^{2}} \\ {\therefore A C=\sqrt{625} \mathrm{cm}=25 \mathrm{cm}}\end{array}\) If sin A = 3/4, Calculate cos A and tan A. 9. NCERT Textbook solutions for class 10 other subjects are also given to free use. \(\begin{array}{l}{\sec \theta=\frac{\text { Hypotenuse }}{\text { Side adjacent to } \angle \theta}} \\ {\frac{13}{12}=\frac{\mathrm{AC}}{\mathrm{AB}}}\end{array}\) Substitute the above equation in the given problem. Applying Pythagoras theorem in \(\triangle ABC \), we obtain Class 10 Maths chapter wise NCERT solution for Maths … Now, apply the values corresponding to the ratios, Now compare the left hand side(LHS) with right hand side(RHS), Hence, (1-tan2 A)/(1+tan2 A) = cos2 A – sin 2 A is proved. Since secant or sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side. So, the statement given above is false. Take advantage of our well-written NCERT solutions for Class 10 Maths Chapter 8 to practice problems related to trigonometry without stress. by cos A. For class 10 CBSE Maths paper, out of the 80 marks, 12 marks are assigned from the unit 5 “Trigonometry”. So, students can download Introduction to Trigonometry Class 10 NCERT Solutions for free and access it any time anywhere. Download NCERT Solutions PDF For Introduction Of Trigonometry Here. In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of: Where k is the positive real number of the problem, Now, substitute the values in the given problem, (i) sin A cos C + cos A sin C = (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1, (ii) cos A cos C – sin A sin C = (√3/2 )(1/2) – (1/2) (√3/2 ) = 0, 10. First divide the numerator and denominator of L.H.S. In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. We will calculate trigonometric ratios for some specific angles and establish some identities involving these ratios, called trigonometric identities. Free PDF of NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry includes all the questions provided in NCERT Books prepared by Mathematics expert teachers as per CBSE NCERT guidelines from Mathongo.com. So, students can download free PDF of Chapter 8 (Introduction to Trigonometry) and take the printout to keep it handy for your exam preparation. You have studied the concept of ratio, in your earlier classes. State whether the following are true or false. (iii) cos A is the abbreviation used for the cosecant of angle A. Justification: Abbreviation used for cosecant of angle M is cosec M. cos M is the abbreviation used for cosine of angle M. Justification: cot M is not the product of cot and M. It is the cotangent of ∠M. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. By following these solutions, you can easily solve and revise the syllabus of CBSE Class 10 Maths. sin 18°/cos 72° = sin (90° – 18°) /cos 72°, we have, Let us assume a right angled triangle ABC, right angled at B. In this chapter, we will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle. Click Here 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), Therefore, (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A.
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