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If tanA=(3)/(4), then show that sin A co...

If `tanA=(3)/(4)`, then show that sin A cos A`=(12)/(25)

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To solve the problem where we need to show that if \( \tan A = \frac{3}{4} \), then \( \sin A \cos A = \frac{12}{25} \), we can follow these steps: ### Step 1: Understand the definition of tangent Given that \( \tan A = \frac{3}{4} \), we can interpret this in terms of a right triangle. The tangent of an angle in a right triangle is defined as the ratio of the opposite side (height) to the adjacent side (base). ### Step 2: Draw a right triangle Let's draw a right triangle where: - The height (opposite side) = 3 ...
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NAGEEN PRAKASHAN-INTRODUCTION TO TRIGONOMETRY-Problems From NCERT/exemplar
  1. If tanA=(3)/(4), then show that sin A cos A=(12)/(25)

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  2. If sqrt(3)tantheta=1 then find value of sin^2theta-cos^2theta

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  3. given : 15 cotA = 8 find sinA and secA

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  4. If s in A=3/4 , calculate cos A and tan A.

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  5. If angleA and angleB are acute angles such that cos A= cos B, then sho...

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  6. (cos45)/(sec30+cosec30)

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  7. Choose the correct option and justify your choice : (i) (2tan30o)/...

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  8. If tan(A+B)=sqrt(3) and t a n(A B)=1/(sqrt(3)); 0o<A+Blt=90o;""...

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  9. In DeltaA B C , right-angled at B, A B"\ "="\ "5"\ "c m and /A C B=30...

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  10. In P Q R , right-angled at Q ,\ P Q=3c m and P R=6c m . Determine ...

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  11. Prove that : sqrt((1-cos^(2)theta)sec^(2)theta)=tantheta

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  12. Prove (tantheta+2)(2tantheta+1)=5tantheta+2sec^2theta.

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  13. If cosA+cos^(2)A=1, then prove that sin^(2)A+sin^(4)A=1.

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  14. Prove the following identity: sec^4theta-sec^2theta=tan^4theta+tan^2th...

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  15. Prove that : 1+(cot^(2)alpha)/(1+"cosec"alpha)="cesec"alpha

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  16. Prove that : (sinalpha+cosalpha)(tanalpha+cotalpha)=secalpha+"cosecα...

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  17. Prove that : (sintheta)/(1+costheta)+(1+costheta)/(sintheta)=2"cosec"t...

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  18. If 2sin^(2)theta-cos^(2)theta=2, then find the value of theta.

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  19. if tantheta+sectheta=l then prove that tantheta=(l^2+1)/(2l)

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  20. If asintheta+bcostheta=c then prove that acostheta-bsintheta=sqrt(a^2+...

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