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R. Lasyapriya

Subject: Maths, asked on 17/5/18

Please answer the following query and please do not send any links.
Q.46. Show that $3 {(\sin θ - \cos θ)}^{4} + 6 {(\sin θ + \cos θ)}^{2} + 4 (\sin^{6} θ + \cos^{6} θ)$ is independent $θ$ .

A.akash

Subject: Maths, asked on 17/5/18

Hi. In RD Sharma book(Pg:11.8) example.4, Can we solve the sum by squaring both sides and get the question rid of square root. Will it is correct?? If we square on both sides, is the eqn is correct???

R. Lasyapriya

Subject: Maths, asked on 16/5/18

Please do not send any links.
$P r o v e : \frac{\cos A}{1 - \sin A} + \frac{\sin A}{1 - \cos A} + 1 = \frac{\sin A \cdot \cos A}{(1 - \sin A) (1 - \cos A)}$

R. Lasyapriya

Subject: Maths, asked on 15/5/18

Solve this : Please solve. Do not send any links. Please answer the question.

$40 . \frac{1}{\cot^{2} θ} + \frac{1}{1 + \tan^{2} θ} = \frac{1}{1 - \sin^{2} θ} - \frac{1}{{cosec}^{2}}$

Suryansh

Subject: Maths, asked on 15/5/18

I'm not able to prove it and have my exam tomorrow. Quick help is highly appreciated:

$Q . \frac{1}{\sec A + \tan A} - \frac{1}{\cos A} = \frac{1}{\cos A} - \frac{1}{\sec A - \tan A}$

Suryansh

Subject: Maths, asked on 15/5/18

Can anyone please solve it?

Example 15 class 10 maths ncert
Example 15: Prove that $\frac{\sin θ - \cos θ + 1}{\sin θ + \cos θ - 1} = \frac{1}{s e c θ - \tan θ}$ , using the identity $s e c^{2} θ = 1 + \tan^{2} θ$ .

R. Lasyapriya

Subject: Maths, asked on 15/5/18

Please answer this query as fast as possible and please dont answer in links solve the answer.

R. Lasyapriya

Subject: Maths, asked on 15/5/18

Solve this : Please solve my queries. Do not send any links.

$Q . Prove that \sec^{2} θ - [\frac{\sin^{2} θ - 2 \sin^{4} θ}{2 \cos^{4} θ - \cos^{2} θ}] = 1$

R. Lasyapriya

Subject: Maths, asked on 15/5/18

Please do not send any links and solve this query . Please answer this query as fast as possible.

$W i t h o u t u \sin g t r i g o n o m e t r i c a l t a b l e e v a l u a t e t h e f o l l o w i n g \frac{\cos 58 °}{\sin 32 °} + \frac{\sin 22 °}{\cos 68 °} - \frac{\cos 38 ° \cos e c 52 °}{\tan 18 ° \tan 35 ° \tan 60 °} - \tan 72 ° \tan 55 °$

R. Lasyapriya

Subject: Maths, asked on 14/5/18

Please answer this question.
Please do not send any links.

$W i t h o u t u \sin g t r i g o n o m e t r i c t a b l e s, e v a l u a t e 2 (\frac{\cos 58 °}{\sin 32 °}) - \sqrt{3} (\frac{\cos 38 ° \cos e c 52 °}{\tan 15 ° \tan 60 ° \tan 75 °})$

R. Lasyapriya

Subject: Maths, asked on 14/5/18

Q.31. Without using trigonometric tables, evaluate the following:
$\frac{s e c 39 °}{\cos e c 51 °} + \frac{2}{\sqrt{3}} \tan 17 ° \tan 38 ° \tan 60 ° \tan 52 ° \tan 73 ° - 3 (\sin^{2} 31 ° + \sin^{2} 59 °)$

R. Lasyapriya

Subject: Maths, asked on 14/5/18

Please answer question 29 .Please don't send any links please.

$29 . E v a l u t e w i t h o u t u \sin g t r i g o n o m e t r i c t a b l e s : \frac{\sin^{2} θ + \sin^{2} (90 - θ)}{3 (s e c^{2} 61 ° - c o t^{2} 29 °)} - \frac{3 c o t^{2} 30 ° \sin^{2} 54 ° s e c^{2} 36 °}{2 (\cos e c^{2} 65 ° - \tan^{2} 25 °)}$ [Foreign 2006]

R. Lasyapriya

Subject: Maths, asked on 13/5/18

Question 151.

Biren Samal

Subject: Maths, asked on 12/5/18

All the trigonometry identities

R. Lasyapriya

Subject: Maths, asked on 10/5/18

151 and 154 Questions .

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